Rd Sharma Xi 2018 Solutions for Class 11 Science Math Chapter 33 Probability are provided here with simple step-by-step explanations. These solutions for Probability are extremely popular among Class 11 Science students for Math Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 11 Science Math Chapter 33 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 11 Science Math are prepared by experts and are 100% accurate.
Page No 33.15:
Question 1:
A coin is tossed. Find the total number of elementary events and also the total number events associated with the random experiment.
Answer:
In tossing a fair coin, there are two possible outcomes, namely Head (H) and Tail (T).
Hence, the sample space in this experiment is given by S = {H, T}.
Thus, total number of elementary events = 2
In all, there are four subsets of S: {H}, {T}, {H, T} and Φ.
Each of the subsets of the sample space is an event.
∴ There are 4 total events associated with the random experiment.
Note: If there are n elements in a set, then the number of its subset is 2^{n}.
Page No 33.15:
Question 2:
List all events associated with the random experiment of tossing of two coins. How many of them are elementary events.
Answer:
The events associated with the random experiment of tossing of two coins are HH, HT, TH and TT. These 4 events are also the elementary events when two unbiased coins are tossed simultaneously.
Page No 33.15:
Question 3:
Three coins are tossed once. Describe the following events associated with this random experiment:
A = Getting three heads
B = Getting two heads and one tail
C = Getting three tails
D = Getting a head on the first coin.
(i) Which pairs of events are mutually exclusive?
(ii) Which events are elementary events?
(iii) Which events are compound events?
Answer:
When three coins are tossed, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Accordingly, we have:
A = {HHH}
B = {HHT, HTH, THH}
C = {TTT}
D = {HHH, HHT, HTH, HTT}
Now, we observe that
A ∩ B = Φ; A ∩ C = Φ; A ∩ D = {HHH} ≠ Φ; B ∩ C = Φ; B ∩ D = {HHT, {HTH} ≠ Φ and C ∩ D = Φ
(i) Events A and B; events A and C; events B and C and events C and D are all mutually exclusive.
(ii) If an event has only one sample point of a sample space, it is called an elementary event.
Thus, A and C are elementary events.
(iii) If an event has more than one sample point of a sample space, it is called a compound event.
Thus, B and D are compound events.
Page No 33.15:
Question 4:
In a single throw of a die describe the following events:
(i) A = Getting a number less than 7
(ii) B = Getting a number greater than 7
(iii) C = Getting a multiple of 3
(iv) D = Getting a number less than 4
(v) E = Getting an even number greater than 4
(vi) F = Getting a number not less than 3.
Also, find A ∪ B, A ∩ B, B ∩ C, E ∩ F, D ∩ F and $\overline{)F}$.
Answer:
When a dice is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6}.
Accordingly, we have:
(i) A = {1, 2, 3, 4, 5, 6}
(ii) B = Φ
(iii) C = {3, 6}
(iv) D = {1, 2, 3}
(v) E = {6}
(vi) F = {3, 4, 5, 6}
Here, A = {1, 2, 3, 4, 5, 6} and B = Φ
∴ A ∪ B = {1, 2, 3, 4, 5, 6}
Here, A = {1, 2, 3, 4, 5, 6} and B = Φ
∴ A ∩ B = Φ
Here, B = Φ and C = {3, 6}
∴ B ∩ C = Φ
Here, E = {6} and F = {3, 4, 5, 6}
∴ E ∩ F = {6}
Here, D = {1, 2, 3} and F = {3, 4, 5, 6}
∴ D ∩ F = {3}
Here, F = {3, 4, 5, 6} and S = {1, 2, 3, 4, 5, 6}
∴ $\overline{\mathrm{F}}=\mathrm{S}-\mathrm{F}=\{1,2\}$
Page No 33.15:
Question 5:
Three coins are tossed. Describe.
(i) two events A and B which are mutually exclusive.
(ii) three events A, B and C which are mutually exclusive and exhaustive.
(iii) two events A and B which are not mutually exclusive.
(iv) two events A and B which are mutually exclusive but not exhaustive.
Answer:
When three coins are tossed, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(i) The two events that are mutually exclusive are as follows:
A: getting no heads
B: getting no tails
This is because sets A = {HHH} and B = {TTT} are disjoint.
(ii) The three events that are mutually exclusive and exhaustive are as follows:
A: getting no heads
B: getting exactly one head
C: getting at least two heads
i.e. A = {TTT}, B = {HTT, THT, TTH} and C = {HHH, HHT, HTH, THH}
This is because A ∩ B = B ∩ C = C ∩ A = Φ and A ∪ B ∪ C = S
(iii) The two events that are not mutually exclusive a
A: getting three heads
B: getting at least 2 heads
i.e. A = {HHH} and B = {HHH, HHT, HTH, THH}
This is because A ∩ B = {HHH} ≠ Φ
(iv) The two events which are mutually exclusive but not exhaustive are as follows:
A: getting exactly one head
B: getting exactly one tail
i.e. A = {HTT, THT, TTH} and B = {HHT, HTH, THH}
It is because, A ∩ B = Φ, but A ∪ B ≠ S
Page No 33.16:
Question 6:
A die is thrown twice. Each time the number appearing on it is recorded. Describe the following events:
(i) A = Both numbers are odd.
(ii) B = Both numbers are even.
(iii) C = sum of the numbers is less than 6
Also, find A ∪ B, A ∩ B, A ∪ C, A ∩ C
Which pairs of events are mutually exclusive?
Answer:
When a dice is thrown twice, we have the following possible outcomes:
A = both numbers are odd
= {(1, 1), (1, 3),(1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
B = both numbers are even
= {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
C = sum of the numbers is less than 6
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
Now, we have:
(A ∪ B) = {(1, 1), (1, 3), (1, 5) ,(3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5),
(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
(A ∩ B) = Φ
(A ∪ C) = {(1, 1), (1, 3),(1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5),
(1, 2), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
(A ∩ C) = {(1, 1), (1, 3), (3, 1)}
Since (A ∩ B) = Φ and (A ∩ C) ≠ Φ, A and B are mutually exclusive, but A and C are not.
Page No 33.16:
Question 7:
Two dice are thrown. The events A, B, C, D, E and F are described as follows:
A = Getting an even number on the first die.
B = Getting an odd number on the first die.
C = Getting at most 5 as sum of the numbers on the two dice.
D = Getting the sum of the numbers on the dice greater than 5 but less than 10.
E = Getting at least 10 as the sum of the numbers on the dice.
F = Getting an odd number on one of the dice.
(i) Describe the following events:
A and B, B or C, B and C, A and E, A or F, A and F
(ii) State true or false:
(a) A and B are mutually exclusive.
(b) A and B are mutually exclusive and exhaustive events.
(c) A and C are mutually exclusive events.
(d) C and D are mutually exclusive and exhaustive events.
(e) C, D and E are mutually exclusive and exhaustive events.
(f) A' and B' are mutually exclusive events.
(g) A, B, F are mutually exclusive and exhaustive events.
Answer:
When two dices are thrown, there are 6^{2} = 36 possible outcomes.
A = Getting an even number on the first dice
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
B = Getting an odd number on the first dice
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),
(3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) }
C = Getting at most 5 as the sum of the numbers on the two dices.
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
D = Getting a sum greater than 5 but less than 10
= {(1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3)}
E = Getting at least 10 as the sum of the numbers on the dices
= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
F = Getting an odd number on one of the dices
= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6),
(4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}
Now,
(i)
A and B = A ∩ B = Φ
B or C = B ∪ C
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3),
(3, 4), (3, 5), (3, 6), (4, 1), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) }
B and C = B ∩ C
= {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}
A and E = A ∩ E
= {(4, 6), (6, 4), (6, 5), (5, 6), (6, 6)}
A or F = A ∪ F
= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2),
(6, 3), (6, 4), (6, 5), (6, 6) }
A and F = A ∩ F
= {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)}
(ii)
(a) True, because A ∩ B = Φ
(b) True, because A ∩ B = Φ and A ∪ B = S
(c) False, because A ∩ C ≠ Φ
(d) False, because C ∩ D = Φ but C ∪ D ≠ S
(e) True, because C ∩ D ∩ E = Φ and C ∪ D ∪ E = S
(f) True, because A' ∩ B' = Φ
(g) False, because A ∩ B ∩ F ≠ Φ and A ∪ B ∪ F = S
Page No 33.16:
Question 8:
The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are then put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the following events:
A = The number on the first slip is larger than the one on the second slip.
B = The number on the second slip is greater than 2
C = The sum of the numbers on the two slips is 6 or 7
D = The number on the second slips is twice that on the first slip.
Which pair(s) of events is (are) mutually exclusive?
Answer:
Four slips marked as 1, 2, 3 and 4 are in a box. Two slips are drawn from it one after the other without replacement. The sample space S for the experiment is
S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}
(i) A = number on the first slip is larger than the one on the second slip
= {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}
(ii) B = number on the second slip is greater than 2
={(1, 3), (2, 3), (1, 4), (2, 4), (3, 4), (4, 3)}
(iii) C = sum of the numbers on the two slips is 6 or 7
={(2, 4), (3, 4), (4, 2), (4, 3)}
(iv) D = number on the second slip is two times the number on the first slip
= {(1, 2), (2, 4)}
Clearly, (A ∩ D) = Φ
Therefore, A and D are mutually exclusive events.
Page No 33.16:
Question 9:
A card is picked up from a deck of 52 playing cards.
(i) What is the sample space of the experiment?
(ii) What is the event that the chosen card is a black faced card?
Answer:
(i) Sample space for a card picked up from a deck of 52 playing card, S
= {A♠, 2♠, 3♠, 4♠, 5♠, 6♠, 7♠, 8♠, 9♠, 10♠, J♠, Q♠, K♠, A♡, 2♡, 3♡, 4♡, 5♡, 6♡, 7♡, 8♡, 9♡, 10♡, J♡, Q♡, K♡, A♣, 2♣, 3♣, 4♣, 5♣, 6♣, 7♣, 8♣, 9♣, 10♣, J♣, Q♣, K♣, A♢, 2♢, 3♢, 4♢, 5♢, 6♢, 7♢, 8♢, 9♢, 10♢, J♢, Q♢, K♢}
(ii) Let A be the event that the chosen card is a black faced card. Then,
A = {J♠, Q♠, K♠, J♣, Q♣, K♣}
Note: ♠ → Spade card, ♡ → Heart card, ♢→ Diamond card, ♣ → Club card
Page No 33.45:
Question 1:
Which of the following cannot be valid assignment of probability for elementary events or outcomes of sample space S = {w_{1}, w_{2}, w_{3}, w_{4}, w_{5}, w_{6}, w_{7}}:
Elementary events: | w_{1} | w_{2} | w_{3} | w_{4} | w_{5} | w_{6} | w_{7} |
(i) | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
(ii) | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ | $\frac{1}{7}$ |
(iii) | 0.7 | 0.6 | 0.5 | 0.4 | 0.3 | 0.2 | 0.1 |
(iv) | $\frac{1}{14}$ | $\frac{2}{14}$ | $\frac{3}{14}$ | $\frac{4}{14}$ | $\frac{5}{14}$ | $\frac{6}{14}$ | $\frac{15}{14}$ |
Answer:
(i)
Here, each of the numbers p(ω_{i}) is positive and less than 1.
∴ Sum of probabilities = $p\left({\omega}_{1}\right)+p\left({\omega}_{2}\right)+p\left({\omega}_{3}\right)+p\left({\omega}_{4}\right)+p\left({\omega}_{5}\right)+p\left({\omega}_{6}\right)+p\left({\omega}_{7}\right)$
= 0. 1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1
Thus, the assignment is valid.
(ii)
Here, each of the numbers p(ω_{i}) is positive and less than 1.
∴ Sum of probabilities = $p\left({\omega}_{1}\right)+p\left({\omega}_{2}\right)+p\left({\omega}_{3}\right)+p\left({\omega}_{4}\right)+p\left({\omega}_{5}\right)+p\left({\omega}_{6}\right)+p\left({\omega}_{7}\right)$
= $\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=7\times \frac{1}{7}=1$
Thus, the assignment is valid.
(iii)
Here, each of the numbers p(ω_{i}) is positive and less than 1.
∴ Sum of probabilities = $p\left({\omega}_{1}\right)+p\left({\omega}_{2}\right)+p\left({\omega}_{3}\right)+p\left({\omega}_{4}\right)+p\left({\omega}_{5}\right)+p\left({\omega}_{6}\right)+p\left({\omega}_{7}\right)$
= 0.7 + 0.6 + 0.5 + 0.4 + 0.3 + 0.2 + 0.1
= 2.8 ≠ 1
Thus, the assignment is not valid.
(iv)
Given:
$p\left({\omega}_{7}\right)=\frac{15}{14}>1$
Thus, the assignment is not valid.
Page No 33.45:
Question 2:
A dice is thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3.
Answer:
The sample space of the given experiment is given by
S = {1, 2, 3, 4, 5, 6}
∴ n (S) = 6
(i) Let A be the event of occurrence of a prime number.
Then A = {2, 3, 5}
i.e. n (A) = 3
$\therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}\mathrm{A}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{\mathrm{n}\left(\mathrm{A}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{3}{6}=\frac{1}{2}$
(ii) Let B be the event of occurrence of the number 2 or 4.
Then B = {2,4}
i.e. n (B) = 2
$\therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}\mathrm{B}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{\mathrm{n}\left(\mathrm{B}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{2}{6}=\frac{1}{3}$
(iii) Let C be the event of occurrence of a multiple of 2 or 3.
Then C = {2, 3, 4, 6}
i.e. n (C) = 4
$\therefore \mathrm{P}\left(\mathrm{C}\right)=\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}\mathrm{C}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{\mathrm{n}\left(\mathrm{C}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{4}{6}=\frac{2}{3}$
Page No 33.45:
Question 3:
In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum more than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) neither a doublet nor a total of 10
(xiii) odd number on the first and 6 on the second
(xiv) a number greater than 4 on each die
(xv) a total of 9 or 11
(xvi) a total greater than 8.
Answer:
We know that in a single throw of two dices, the total number of possible outcomes is (6 × 6) = 36.
Let S be the sample space.
Then n(S) = 36
(i) Let E_{1} = event of getting 8 as the sum.
Then E_{1} = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
i.e. n (E_{1}) = 5
$\therefore P\left({E}_{1}\right)=\frac{n\left({E}_{1}\right)}{n\left(S\right)}=\frac{5}{36}$
(ii) Let E_{2} = event of getting a doublet
Then E_{2} = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
i.e. n (E_{2}) = 6
$\therefore P\left({E}_{2}\right)=\frac{n\left({E}_{2}\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$
(iii) Let E_{3} = event of getting a doublet of prime numbers
Then E_{3} = {(2, 2), (3, 3), (5, 5)}
i.e. n (E_{3}) = 3
$\therefore P\left({E}_{3}\right)=\frac{n\left({E}_{3}\right)}{n\left(S\right)}=\frac{3}{36}=\frac{1}{12}$
(iv) Let E_{4} = event of getting a doublet of odd numbers
Then E_{4} = {(1, 1), (3, 3), (5, 5)}
i.e. n (E_{4}) = 3
$\therefore P\left({E}_{4}\right)=\frac{n\left({E}_{4}\right)}{n\left(S\right)}=\frac{3}{36}=\frac{1}{12}$
(v) Let E_{5} = event of getting a sum greater than 9
Then E_{5} = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
i.e. n (E_{5}) = 6
$\therefore P\left({E}_{5}\right)=\frac{n\left({E}_{5}\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$
(vi) Let E_{6} = event of getting an even number on the first throw
Then E_{6} = {(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1) , (6, 2), (6, 3),
(6, 4), (6, 5), (6, 6) }
i.e. n (E_{6}) = 18
$\therefore P\left({E}_{6}\right)=\frac{n\left({E}_{6}\right)}{n\left(S\right)}=\frac{18}{36}=\frac{1}{2}$
(vii) Let E_{7} = event of getting an even number on one dice and a multiple of 3 on the other
Then E_{7} = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6) , (3, 2), (6, 2), (3, 4), (6, 4), (3, 6)}
i.e. n (E_{7}) = 11
$\therefore P\left({E}_{7}\right)=\frac{n\left({E}_{7}\right)}{n\left(S\right)}=\frac{11}{36}$
(viii) Let E_{8} = event of getting neither 9 nor 11 as the sum of the numbers on the faces
Then $\overline{{E}_{8}}$ = event of getting either 9 or 11 as the sum
Thus, $\overline{){E}_{8}}$ = {(3, 6), (4, 5), (5, 4) , (5, 6), (6, 3), (6, 5) }
$\mathrm{i}.\mathrm{e}.n\left(\overline{{E}_{8}}\right)=6$
$\therefore P\left(\overline{{E}_{8}}\right)=\frac{n\left(\overline{{E}_{8}}\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$
$\mathrm{Hence},P\left({E}_{8}\right)=1-P\left(\overline{{E}_{8}}\right)$
$=1-\frac{1}{6}=\frac{5}{6}$
(ix) Let E_{9} = event of getting a sum less than 6
Then E_{9} = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
i.e. n (E_{9}) = 10
$\therefore P\left({E}_{9}\right)=\frac{n\left({E}_{9}\right)}{n\left(S\right)}=\frac{10}{36}=\frac{5}{18}$
(x) Let E_{10}_{ }= event of getting a sum less than 7
Then E_{10} = {(1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (2, 1) , (2, 2), (2, 3), (2, 4), (3, 1) , (3, 2), (3, 3), (4, 1) , (4, 2), (5, 1)}
i.e. n (E_{10}) = 15
$\therefore P\left({E}_{10}\right)=\frac{n\left({E}_{10}\right)}{n\left(S\right)}=\frac{15}{36}=\frac{5}{12}$
(xi) Let E_{11} = event of getting a sum greater than 7
Then E_{11} = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
i.e. n (E_{11}) = 15
$\therefore P\left({E}_{11}\right)=\frac{n\left({E}_{11}\right)}{n\left(S\right)}=\frac{15}{36}=\frac{5}{12}$
(xii) Let E_{12} = event of getting neither a doublet nor a total of 10
Thus $\overline{{E}_{12}}$ = event of getting either a doublet or a total of 10
Then $\overline{{E}_{12}}$ = {(1, 1), (2, 2), (3, 3), (4, 4), (4, 6), (5, 5), (6, 4), (6, 6)}
$\mathrm{i}.\mathrm{e}.n\left(\overline{{E}_{12}}\right)=8$
$\therefore P\overline{\left({E}_{12}\right)}=\frac{n\overline{\left({E}_{12}\right)}}{n\left(S\right)}=\frac{8}{36}=\frac{2}{9}$
$\mathrm{Hence},P\left({E}_{12}\right)=1-P\left(\overline{{E}_{12}}\right)\phantom{\rule{0ex}{0ex}}$
$=1-\frac{2}{9}=\frac{7}{9}$
(xiii) Let E_{13} = event of getting an odd number on the first throw and 6 on the second
Then E_{13} = {(1,6), (3, 6), (5, 6)}
i.e. n (E_{13}) = 3
$\therefore P\left({E}_{13}\right)=\frac{n\left({E}_{13}\right)}{n\left(S\right)}=\frac{3}{36}=\frac{1}{12}$
(xiv) Let E_{14} = event of getting a number greater than 4 on each dice
Then E_{14} = {(5, 5), (5, 6), (6, 5), (6, 6)}
i.e. n (E_{14}) = 4
$\therefore P\left({E}_{14}\right)=\frac{n\left({E}_{14}\right)}{n\left(S\right)}=\frac{4}{36}=\frac{1}{9}$
(xv) Let E_{15} = event of getting a total of 9 or 11
Then E_{15} = {(3, 6), (4, 5), (5, 4) , (5, 6), (6, 3), (6, 5) }
i.e. n (E_{15}) = 6
$\therefore P\left({E}_{15}\right)=\frac{n\left({E}_{15}\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$
(xvi) Let E_{16} = event of getting a total greater than 8
Then E_{16} = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}
i.e. n (E_{16}) = 10
$\therefore P\left({E}_{16}\right)=\frac{n\left({E}_{16}\right)}{n\left(S\right)}=\frac{10}{36}=\frac{5}{18}$
Page No 33.46:
Question 4:
In a single throw of three dice, find the probability of getting a total of 17 or 18.
Answer:
If three dices are thrown simultaneously, then all the possible outcomes = 6^{3} = 216
∴ Total number of possible outcome, n(S) = 216
Let A = event of getting a sum of numbers on three dice as 17 or 18
Then the favourable outcomes are given as
A = {(6, 6, 5), (6, 5, 6), (5, 6, 6), (6, 6, 6)}
Number of favourable outcomes, n(A) = 4
Hence, required probability, P(A) = P (sum of the numbers on three dices as 17 or 18) = $\frac{4}{216}=\frac{1}{54}$
Page No 33.46:
Question 5:
Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail.
Answer:
When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
i.e. n (S) = 8
(i) Let E_{1} = event of getting exactly two heads
Then E_{1} = {HHT, HTH, THH}
i.e. n(E_{1}) = 3
$\therefore P\left({E}_{1}\right)=\frac{n\left({E}_{1}\right)}{n\left(S\right)}=\frac{3}{8}$
(ii) Let E_{2} = event of getting at least two heads
Then E_{2} = {HHH, HHT, HTH, THH}
i.e. n (E_{2}) = 4
$\therefore P\left({E}_{2}\right)=\frac{n\left({E}_{2}\right)}{n\left(S\right)}=\frac{4}{8}=\frac{1}{2}$
(iii) Let E_{3} = event of getting at least one head and one tail
Then E_{3} = {HHT, HTH, HTT, THH, THT, TTH}
i.e. n(E_{3}) = 6
$\therefore P\left({E}_{3}\right)=\frac{n\left({E}_{3}\right)}{n\left(S\right)}=\frac{6}{8}=\frac{3}{4}$
Page No 33.46:
Question 6:
What is the probability that an ordinary year has 53 Sundays?
Answer:
There are 365 days, i.e. 52 weeks and one day
This one day can be any of the seven days of the week.
∴ P (Sunday) = $\frac{1}{7}$
We know that 52 weeks will have 52 Sundays.
Hence, required probability = P (an ordinary year with 53 Sundays) = $\frac{1}{7}$
Page No 33.46:
Question 7:
What is the probability that a leap year has 53 Sundays and 53 Mondays?
Answer:
We know that a leap year has 366 days (i.e. 7 $\times $ 52 + 2) = 52 weeks and 2 extra days.
The sample space for these two extra days is given by
S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
There are 7 cases.
i.e. n(S) = 7
Let E be the event in which the leap year has 53 Sundays and 53 Mondays.
Then E = {(Sunday, Monday) }
i.e. n(E) = 1
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{1}{7}$
Hence, the probability in which a leap year has 53 Sundays and 53 Mondays is $\frac{1}{7}.$
Page No 33.46:
Question 8:
A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that:
(i) All the three balls are white.
(ii) All the three balls are red.
(iii) One ball is red and two balls are white.
Answer:
Total number of balls = 8 + 5 = 13
Total number of events for drawing 3 balls = ${}^{13}{C}_{3}$
(i) Total number of events for getting white balls = ${}^{5}{C}_{3}$
P(all 3 balls white) = $\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{outcomes}}={\frac{{}^{5}{\displaystyle {\mathrm{C}}_{3}}}{{}^{13}{\displaystyle \mathrm{C}}}}_{3}$
$=\frac{5!}{3!\times 2!}\times \frac{3!\times 10!}{13!}$
$\Rightarrow $P(all 3 balls white)$=\frac{5}{143}$
(ii) Favourable number of events for getting red balls = ${}^{8}{C}_{3}$
P(all 3 balls red) = $\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{outcomes}}={\frac{{}^{8}{\displaystyle {\mathrm{C}}_{3}}}{{}^{13}{\displaystyle \mathrm{C}}}}_{3}$
$=\frac{8!}{3!\times 5!}\times \frac{3!\times 10!}{13!}$
$\Rightarrow $P(all 3 balls red)$=\frac{28}{143}$
(iii) Favourable number of events for getting 1 red ball = ${}^{8}{C}_{1}$
Favourable number of events for getting 2 white balls = ${}^{5}{C}_{2}$
P(1 red and 2 white balls) = $\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{outcomes}}=\frac{{}^{8}{\displaystyle {\mathrm{C}}_{1}}{\times}^{5}{C}_{2}}{{}^{13}{\displaystyle {\mathrm{C}}_{3}}}$
$=\frac{8!}{1!\times 7!}\times \frac{5!}{2!\times 3!}\times \frac{3!\times 10!}{13!}$
$\Rightarrow $P(all 3 balls red)$=\frac{40}{143}$
Page No 33.46:
Question 9:
In a single throw of three dice, find the probability of getting the same number on all the three dice.
Answer:
If three dices are thrown simultaneously, then all the possible outcomes = 6^{3} = 216
∴ Total number of possible outcome, n(S) = 216
Let A be the event of getting the same number on all the three dices.
Then, the favourable outcomes are as follows:
A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6 , 6)}
Number of favourable outcomes, n(A) = 6
Hence, required probability, P(A) = P (same number on all the three dices) = $\frac{6}{216}=\frac{1}{36}$
Page No 33.46:
Question 10:
Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.
Answer:
If two dices are thrown simultaneously, then all the possible outcomes = 6^{2} = 36
∴ Total number of possible outcome, n(S) = 36
Let A = event where the the total of the numbers on the dices is greater than 10
Then the favourable outcomes are as follows:
A = {(5, 6), (6, 5), (6, 6)}
Number of favourable outcomes, n(A) = 3
Hence, required probability, P(A) = P (total of the numbers on the dices is greater than 10) = $\frac{3}{36}=\frac{1}{12}$
Page No 33.46:
Question 11:
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither a heart nor a king
(vi) spade or an ace
(vii) neither an ace nor a king
(viii) a diamond card
(ix) not a diamond card
(x) a black card
(xi) not an ace
(xii) not a black card.
Answer:
Let S denote the sample space.
Then, n(S) = 52
(i) Let E_{1} = event of drawing a black king
We know that the number of black kings is two: one for spade and one for club.
i.e. n (E_{1}) = 2
$\therefore P\left({E}_{1}\right)=\frac{n\left({E}_{1}\right)}{n\left(S\right)}=\frac{2}{52}=\frac{1}{26}$
(ii) Let E_{2} = event of drawing either a black card or a king
There are 26 black cards including two black kings and there are two more kings.
Therefore, there are 28 cards in which there is a black card or a king.
i.e. n (E_{2}) = 28
$\therefore P\left({E}_{2}\right)=\frac{n\left({E}_{2}\right)}{n\left(S\right)}=\frac{28}{52}=\frac{7}{13}$
(iii) Let E_{3} = event of drawing black and a king
There are two cards: a black card and a king, i.e. a black king.
i.e. n (E_{3}) = ^{2}C_{1} = 2
$\therefore P\left({E}_{3}\right)=\frac{n\left({E}_{3}\right)}{n\left(S\right)}=\frac{2}{52}=\frac{1}{26}$
(iv) Let E_{4} = event of drawing a jack, a queen or a king
Out of 52 cards, there are four jacks, four queens and four kings.
i.e. n (E_{4}) = ^{4}C_{1} + ^{4}C_{1} + ^{4}C_{1}
= 4 + 4 + 4 = 12
$\therefore P\left({E}_{4}\right)=\frac{n\left({E}_{4}\right)}{n\left(S\right)}=\frac{12}{52}=\frac{3}{13}$
(v) Let E_{5} = event of drawing neither a heart nor a king
Then $\overline{{E}_{5}}$ = event of drawing either a heart or a king
There are 13 cards of heart including one king. Also, there are 3 more kings.
Therefore, out of these 16 cards, one can draw either a heart or a king in ^{16}C_{1} ways.
$\mathrm{i}.\mathrm{e}.n\left(\overline{{E}_{5}}\right)=16$
$\therefore P\left(\overline{{E}_{5}}\right)=\frac{n\left(\overline{{E}_{5}}\right)}{n\left(S\right)}=\frac{16}{52}=\frac{4}{13}$
$\therefore P\left({E}_{5}\right)=1-P\left(\overline{{E}_{5}}\right)\phantom{\rule{0ex}{0ex}}$
$=1-\frac{4}{13}=\frac{9}{13}$
(vi) Let E_{6} = event of drawing a spade or an ace
There are 13 spade cards including one ace. Also, there are 3 more ace cards.
Therefore, out of these 16 cards, one can draw either a spade or an ace in ^{16}C_{1} ways.
i.e. n (E_{6}) = 16
∴ $P\left({E}_{6}\right)=\frac{n\left({E}_{6}\right)}{n\left(S\right)}=\frac{16}{52}=\frac{4}{13}$
(vii) Let E_{7} = event of drawing neither an ace nor a king
Then $\overline{{E}_{7}}$ = event of drawing either an ace or a king
There are four ace cards and four king cards.
Therefore, out of these 8 cards, one can draw either an ace or a king in ^{8}C_{1} ways.
$\mathrm{i}.\mathrm{e}.n\left(\overline{{E}_{7}}\right)=8$
$\therefore P\left(\overline{{E}_{7}}\right)=\frac{n\left(\overline{{E}_{7}}\right)}{n\left(S\right)}=\frac{8}{52}=\frac{2}{13}$
$\therefore P\left({E}_{7}\right)=1-P\left(\overline{{E}_{7}}\right)\phantom{\rule{0ex}{0ex}}$
$=1-\frac{2}{13}=\frac{11}{13}$
(viii) Let E_{8} = event of drawing a diamond card.
There are 13 diamond cards in a pack of 52 cards, out of which one diamond can be drawn in ^{13}C_{1} .
∴ n (E_{8}) = 13
$P\left({E}_{8}\right)=\frac{n\left({E}_{8}\right)}{n\left(S\right)}=\frac{13}{52}=\frac{1}{4}$
(ix) Let E_{9} = event of not drawing a diamond card
Then $\overline{{E}_{9}}$ = event of drawing a diamond card
There are 13 diamond cards in a pack of 52 cards, out of which one diamond card can be drawn in ^{13}C_{1}.
$\mathrm{i}.\mathrm{e}.n\left(\overline{{E}_{9}}\right)=\frac{13}{52}=\frac{1}{4}$
$\therefore P\left({E}_{9}\right)=1-P\left(\overline{){E}_{9}}\right)\phantom{\rule{0ex}{0ex}}$
$=1-\frac{1}{4}=\frac{3}{4}$
(x) Let E_{10}_{ }= event of drawing a black card
We know that there are 26 black cards, i.e. 13 spades and 13 clubs.
Then n (E_{10}) = 26
$\therefore P\left({E}_{10}\right)=\frac{n\left({E}_{10}\right)}{n\left(S\right)}=\frac{26}{52}=\frac{1}{2}$
(xi) Let E_{11} = event of drawing a card which is not an ace
Then $\overline{{E}_{11}}$ = event of drawing an ace card
There are four aces in a pack of 52 cards, out of which one ace can be drawn in ^{4}C_{1} ways.
$\mathrm{i}.\mathrm{e}.n\left(\overline{){E}_{11}}\right)=4$
$\therefore P\left(\overline{){E}_{11}}\right)=\frac{n\left(\overline{){E}_{11}}\right)}{n\left(S\right)}=\frac{1}{13}$
$\mathrm{Hence},P\left({E}_{11}\right)=1-P\left(\overline{){E}_{11}}\right)\phantom{\rule{0ex}{0ex}}$
$=1-\frac{1}{13}=\frac{12}{13}$
(xii)
Let E_{12} = event of drawing a non-black card
We know that there are 26 non-black cards, i.e. 13 diamonds and 13 hearts.
Then n (E_{12}) = 26
$\therefore P\left({E}_{12}\right)=\frac{n\left({E}_{12}\right)}{n\left(S\right)}=\frac{26}{52}=\frac{1}{2}$
Page No 33.46:
Question 12:
In shuffling a pack of 52 playing cards, four are accidently dropped; find the chance that the missing cards should be one from each suit.
Answer:
It is given that from a well-shuffled pack of cards, four cards are missing out.
∴ Total number of elementary events, n(S) = ^{52}C_{4}
Let E be the event where four cards are missing from each suit.
i.e. n(E)= ^{13}C_{1} × ^{13}C_{1} × ^{13}C_{1} × ^{13}C_{1}
Hence, required probability = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{{}^{13}C_{1}\times {}^{13}C_{1}\times {}^{13}C_{1}\times {}^{13}C_{1}}{{}^{52}C_{4}}\phantom{\rule{0ex}{0ex}}$
$=\frac{13\times 13\times 13\times 13}{{\displaystyle \frac{52\times 51\times 50\times 49}{4\times 3\times 2\times 1}}}\phantom{\rule{0ex}{0ex}}=\frac{13\times 13\times 13\times 13}{13\times 17\times 25\times 49}\phantom{\rule{0ex}{0ex}}=\frac{2197}{20825}$
Page No 33.46:
Question 13:
From a deck of 52 cards, four cards are drawn simultaneously, find the chance that they will be the four honours of the same suit.
Answer:
Four cards can be drawn from a pack of 52 cards in ^{52}C_{4} ways.
i.e. n(S) = ^{52}C_{4}
Of these, there are four ways to draw all the four honours of a suit, i.e. J, Q, K and A of Spades, Hearts, Diamonds or Clubs.
Let E be the favourable event of that all the four cards drawn are honour cards from the same suit.
Favourable number of events, n(E) = ^{4}C_{4} or ^{4}C_{4}_{ }or ^{4}C_{4}_{ }or ^{4}C_{4}_{ }= ^{4}C_{4}_{ }+ ^{4}C_{4}_{ }+ ^{4}C_{4}_{ }+ ^{4}C_{4}_{ }
Hence, required probability = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{{}^{4}C_{4}+{}^{4}C_{4}+{}^{4}C_{4}+{}^{4}C_{4}}{{}^{52}C_{4}}\phantom{\rule{0ex}{0ex}}$
$=\frac{4}{{\displaystyle \frac{52\times 51\times 50\times 49}{4\times 3\times 2\times 1}}}\phantom{\rule{0ex}{0ex}}=\frac{4}{13\times 17\times 25\times 49}\phantom{\rule{0ex}{0ex}}=\frac{4}{270725}$
Page No 33.46:
Question 14:
Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?
Answer:
Total number of elementary events, n(S) = ^{20}C_{1} = 20
Multiples of 3 or 7 = 3, 6, 9, 12, 15, 18, 7, 14
Thus, favourable number of events, n(E) = ^{8}C_{1} = 8
Hence, required probability = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{8}{20}=\frac{2}{5}$
Page No 33.46:
Question 15:
A bag contains 6 red, 4 white and 8 blue balls. if three balls are drawn at random, find the probability that one is red, one is white and one is blue.
Answer:
Total number of balls = 6 + 4 + 8 = 18
Total number of elementary events, n(S) = ^{18}C_{3}
Let E be the event of favourable outcomes.
Here, E = getting one red, one white and one blue ball
So, favourable number of elementary events, n(E) = ^{6}C_{1} ×^{4}C_{1} ×^{ }^{8}C_{1}
Hence, required probability = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{{}^{6}C_{1}\times {}^{4}C_{1}\times {}^{8}C_{1}}{{}^{18}C_{3}}\phantom{\rule{0ex}{0ex}}$
$=\frac{6\times 4\times 8}{{\displaystyle \frac{18\times 17\times 16}{3\times 2}}}\phantom{\rule{0ex}{0ex}}=\frac{6\times 4\times 8}{6\times 17\times 8}\phantom{\rule{0ex}{0ex}}=\frac{4}{17}$
Page No 33.46:
Question 16:
A bag contains 7 white, 5 black and 4 red balls. If two balls are drawn at random, find the probability that:
(i) both the balls are white
(ii) one ball is black and the other red
(iii) both the balls are of the same colour.
Answer:
Out of 16 balls, two balls can be drawn in ^{16}C_{2} ways.
∴ Total number of elementary events = ^{16}C_{2} = 120
(i)
Out of seven white balls, two white balls can be chosen in ^{7}C_{2} ways.
∴ Favourable number of ways = ^{7}C_{2}
Hence, required probability = $\frac{{}^{7}C_{2}}{{}^{16}C_{2}}=\frac{21}{120}=\frac{7}{40}$
(ii)
Out of five black balls, one black ball can be drawn in ^{5}C_{1} ways.
Out of four red balls, one red ball can be drawn in ^{4}C_{1} ways.
Therefore, one black and one red balls can be drawn in ^{5}C_{1}× ^{4}C_{1} ways.
∴ Favourable number of ways = ^{5}C_{1}× ^{4}C_{1} = 5× 4 = 20
Hence, required probability = $\frac{20}{120}=\frac{1}{6}$
(iii)
'Two balls drawn are of the same colour' means that both are either white or black or red.
Out of seven white balls, two white balls can be drawn in ^{7}C_{2} ways.
Similarly, two black balls can be drawn from five black balls in ^{5}C_{2} ways and two red balls can be drawn from four red balls in ^{4}C_{2} ways.
Therefore, number of ways of drawing two balls of the same colour = ^{7}C_{2} + ^{5}C_{2} + ^{4}C_{2} = 21 + 10 + 6 = 37
i.e. favourable number of ways = 37
Hence, required probability = $\frac{37}{120}$
Page No 33.46:
Question 17:
A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that:
(i) one is red and two are white
(ii) two are blue and one is red
(iii) one is red
Answer:
Out of 18 balls, three balls can be drawn in ^{18}C_{3} ways.
∴ Total number of elementary events = ^{18}C_{3} = 816
(i)
Out of six red balls, one red ball can be drawn in ^{6}C_{1} ways.
Out of four white balls, two white balls can be drawn in ^{4}C_{2} ways .
Therefore, one red and two white balls can be drawn in ^{6}C_{1} × ^{4}C_{2}_{ }= 6 × 6 = 36 ways
∴ Favourable number of ways = 36
Hence, required probability = $\frac{36}{816}=\frac{3}{68}$
(ii)
Out of eight blue balls, two blue balls can be drawn in ^{8}C_{2} ways.
Out of six red balls, one red ball can be drawn in ^{6}C_{1} ways .
Therefore, two blue and one red balls can be drawn in ^{8}C_{2} × ^{6}C_{1} = 28 × 6 = 168 ways
∴ Favourable number of ways = 168
Hence, required probability = $\frac{168}{816}=\frac{7}{34}$
(iii)
There are six red balls out of which one red ball can be drawn in ^{6}C_{1} ways.
Two balls from the remaining 12 balls can be drawn in ^{12}C_{2} ways.
Therefore, one red two other coloured balls can be drawn in ^{6}C_{1}× ^{12}C_{2} = 6 × 66 = 396 ways
∴ Favourable number of ways = 396
Hence, required probability = $\frac{396}{816}=\frac{33}{68}$
Page No 33.46:
Question 18:
Five cards are drawn from a pack of 52 cards. What is the chance that these 5 will contain:
(i) just one ace
(ii) at least one ace?
Answer:
Let S denote the sample space.
Then n(S) = 52
Thus, five cards can be drawn in ^{52}C_{5} ways.
∴ Total number of elementary events = ^{52}C_{5}
(i)
Let E_{1} = event of getting just one ace
i.e. E_{1} = ^{4}C_{1} × ^{48}C_{4}
∴ Total number of favourable events = ^{4}C_{1} × ^{48}C_{4}
Hence, required probability = $\frac{{}^{4}C_{1}\times {}^{48}C_{4}}{{}^{52}C_{5}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$=\frac{4\times 2\times 47\times 46\times 45}{52\times 51\times 10\times 49\times 2}\phantom{\rule{0ex}{0ex}}$
$=\frac{47\times 46\times 9}{13\times 51\times 2\times 49}$
$=\frac{47\times 23\times 3}{13\times 17\times 49}=\frac{3243}{10829}$
(ii)
Probability for at least one ace = 1 – Probability (no ace)
= $1-\frac{{}^{48}C_{5}}{{}^{52}C_{5}}$
= $1-\frac{35673}{54145}=\frac{18472}{54145}$
Page No 33.46:
Question 19:
The face cards are removed from a full pack. Out of the remaining 40 cards, 4 are drawn at random. what is the probability that they belong to different suits?
Answer:
Having removed 12 face cards, the remaining 40 cards include 10 cards in each suit.
∴ Chance of drawing a card in the first draw = $\frac{{}^{40}C_{1}}{{}^{40}C_{1}}=1$
Having drawn 1 card, there remain 39 cards of which 30 are of suits different from the drawn card.
∴ Chance of drawing a card of different suit in the second draw = $\frac{{}^{30}C_{1}}{{}^{39}C_{1}}=\frac{30}{39}$
Having drawn two cards, there remain 38 cards of which 20 are of suits different from the drawn cards.
∴ Chance of drawing a card of in the third draw = $\frac{{}^{20}C_{1}}{{}^{38}C_{1}}=\frac{20}{38}=\frac{10}{19}$
Having drawn three cards, there remain 37 cards of which 10 are of suits different from the drawn cards.
∴ Chance of drawing a card in the fourth draw = $\frac{{}^{10}C_{1}}{{}^{37}C_{1}}=\frac{10}{37}$
Hence, all the events being dependent, the required probability = $1\times \frac{30}{39}\times \frac{10}{19}\times \frac{10}{37}=\frac{1000}{9139}$
Page No 33.46:
Question 20:
There are four men and six women on the city councils. If one council member is selected for a committee at random, how likely is that it is a women?
Answer:
Out of four men and six women, one person can be chosen in ^{10}C_{1} = 10 ways.
Number of ways of selecting one women out of six women = ^{6}C_{1} = 6 ways.
$\therefore \mathrm{Required}\mathrm{probability}=\frac{6}{10}=\frac{3}{5}$
Page No 33.47:
Question 21:
A box contains 100 bulbs, 20 of which are defective. 10 bulbs are selected for inspection. Find the probability that:
(i) all 10 are defective
(ii) all 10 are good
(iii) at least one is defective
(iv) none is defective
Answer:
Out of 100 bulbs, 10 can be chosen in ^{100}C_{10} ways.
So, total number of elementary events = ^{100}C_{10}
(i)
There are 20 defective and 80 non-defective bulbs.
The number of ways of selecting 10 defective bulbs out of 20 is ^{20}C_{10} ways.
∴ Favourable number of elementary events = ^{20}C_{10} ways
Hence, required probability = $\frac{{}^{20}C_{10}}{{}^{100}C_{10}}$
(ii)
The number of ways of selecting 10 non-defective bulb out of 80 is ^{80}C_{10} ways.
∴ Favourable number of elementary events = ^{80}C_{10}
Hence, required probability = $\frac{{}^{80}C_{10}}{{}^{100}C_{10}}$
(iii)
Probability for at least one defective bulb = 1 – Probability (all 10 are non-defective)
= $1-\frac{{}^{80}C_{10}}{{}^{100}C_{10}}$
(iv)
'None is defective' means that all are non-defective bulbs. The number of ways of selecting all 10 non-defective bulbs out
of 80 is ^{80}C_{10} ways.
∴ Favourable number of elementary events = ^{80}C_{10}
Hence, required probability = $\frac{{}^{80}C_{10}}{{}^{100}C_{10}}$
Page No 33.47:
Question 22:
Find the probability that in a random arrangement of the letters of the word 'SOCIAL' vowels come together.
Answer:
There are six letters in the word ‘SOCIAL’, which can be arranged in 6! ways.
There are three vowels, namely O, I and A.
Let us consider these three vowels as one letter.
So, when the three vowels are clubbed together, we have (O, I, A) SCL. We can arrange four letters in a row in 4! ways.
Also, the three vowels can themselves be arranged in 3! ways.
Hence, required probability = $\frac{4!\times 3!}{6!}=\frac{4!\times 3\times 2}{6\times 5\times 4!}=\frac{1}{5}$
Page No 33.47:
Question 23:
The letters of the word' CLIFTON' are placed at random in a row. What is the chance that two vowels come together?
Answer:
There are 7 letters in the word ‘CLIFTON’, which can be arranged in 7! ways.
There are two vowels, namely I and O.
Let us consider these two vowels as one letter.
So, when the two vowels are clubbed together, we have (I,O) CLFTN.
We can arrange six letters in a row in 6! ways.
Also, the two vowels can be arranged in 2! ways.
Hence, required probability = $\frac{6!\times 2!}{7!}=\frac{6!\times 2}{7\times 6!}=\frac{2}{7}$
Page No 33.47:
Question 24:
The letters of the word 'FORTUNATES' are arranged at random in a row. What is the chance that the two 'T' come together.
Answer:
There are 10 letters in the word ‘FORTUNATES’, which can be arranged in 10! ways.
There are two T's in the word.
Let us consider these two letters in the word ‘FORTUNATES’ as one letter.
So, when the two T's are clubbed together, we have (T,T) FORUNAES.
We can arrange 9 letters in a row in 9! ways.
Also, the two T's can themselves be arranged in 2! ways.
Hence, required probability = $\frac{9!\times 2!}{10!}=\frac{9!\times 2}{10\times 9!}=\frac{1}{5}$
Page No 33.47:
Question 25:
A committee of two persons is selected from two men and two women. What is the probability that the committee will have (i) no man? (ii) one man? (iii) two men?
Answer:
Total number of people = 2 + 2 = 4
Out of these four people, two can be selected in ^{4}C_{2} = 6 ways.
(i) No man in the committee of two means that there will be two women in the committee.
Out of two women, two can be selected in ^{2}C_{2} = 1 way.
∴ P(no man) = $\frac{{}^{2}\mathrm{C}_{2}}{{}^{4}\mathrm{C}_{2}}=\frac{1}{2\times 3}=\frac{1}{6}$
(ii) One man in the committee of two means that there is one woman in the committee.
One man out of 2 can be selected in ^{2}C_{1} = 2 ways.
One woman out of 2 can be selected in ^{2}C_{1} = 2 ways.
Together, they can be selected in ^{2}C_{1} × ^{2}C_{1} ways.
∴ P (one man) = $\frac{{}^{2}\mathrm{C}_{1}\times {}^{2}\mathrm{C}_{1}}{{}^{4}\mathrm{C}_{2}}=\frac{2\times 2}{2\times 3}=\frac{2}{3}$
(iii) Two men can be selected in ^{2}C_{2} way.
∴ P (two men) = $\frac{{}^{2}\mathrm{C}_{2}}{{}^{4}\mathrm{C}_{2}}=\frac{1}{2\times 3}=\frac{1}{6}$
Page No 33.47:
Question 26:
If odds in favour of an event be 2 : 3, find the probability of occurrence of this event.
Answer:
If odds in favour of an event A are a : b, then probability of happening of event A = P(A) = $\frac{a}{a+b}$
It is given that the odds in favour of an event are 2 : 3.
Thus, n(S) = 2k + 3k = 5k and n(E) = 2k
Hence, probability of occurrence of this event = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{2k}{5k}=\frac{2}{5}$
Page No 33.47:
Question 27:
If odds against an event be 7 : 9, find the probability of non-occurrence of this event.
Answer:
If the odds against the happening of an event A are a : b, then the probability of not happening of event A = $P\left(\overline{A}\right)=\frac{a}{a+b}$
It is given that odds against an event are 7 : 9.
Thus, n(S) = 7k + 9k = 16k
and n(E) = 7k
Hence, probability of non-occurrence of this event = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{7k}{16k}=\frac{7}{16}$
Page No 33.47:
Question 28:
Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one by one without, replacement. Find the probability that both the balls are of different colours.
Answer:
Out of 14 balls, two balls can be chosen in ^{14}C_{2} ways.
So, favourable number of elementary events = ^{14}C_{2}
Let A be the event of getting two balls of different colours.
∴ P(A) = P(1 white and 1 red) + P(1 white and 1 green) + P(1 white and 1 black) + P(1 red and 1 green)
+ P(1 red and 1 black) + P(1 green and 1 black)
$\Rightarrow P\left(A\right)=\frac{C\left(2,1\right)C\left(3,1\right)}{C\left(14,2\right)}+\frac{C\left(2,1\right)C\left(5,1\right)}{C\left(14,2\right)}+\frac{C\left(2,1\right)C\left(4,1\right)}{C\left(14,2\right)}+\frac{C\left(3,1\right)C\left(5,1\right)}{C\left(14,2\right)}+\frac{C\left(3,1\right)C\left(4,1\right)}{C\left(14,2\right)}+\frac{C\left(5,1\right)C\left(4,1\right)}{C\left(14,2\right)}$
$=\frac{2\times 3}{91}+\frac{2\times 5}{91}+\frac{2\times 4}{91}+\frac{3\times 5}{91}+\frac{3\times 4}{91}+\frac{5\times 4}{91}\phantom{\rule{0ex}{0ex}}$
$=\frac{6+10+8+15+12+20}{91}=\frac{71}{91}=0.78$
Page No 33.47:
Question 29:
Two unbiased dice are thrown. Find the probability that:
(i) neither a doublet nor a total of 8 will appear
(ii) the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3
Answer:
We know that in a single throw of two dices, the total number of possible outcomes is (6 × 6) = 36.
Let S be the sample space.
Then n(S) = 36
(i)
Let E_{1} = event of getting neither a doublet nor a total of 8
Then E_{1}' = event of getting either a doublet or a total of 8
∴ E_{1}' = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (2, 6), (3, 5), (5, 3), (6, 2)}
i.e. n(E_{1}') = 10
Thus, P(E_{1}') = $\frac{10}{36}$
Hence, required probability P(E_{1}) = 1 $-$ P(E_{1}')
$=1-\frac{10}{36}=\frac{36-10}{36}=\frac{26}{36}=\frac{13}{18}$
(ii)
Let E_{2} be the event of getting the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3. Then,
E_{2}' = event of getting the sum of the numbers obtained on the two dice is either a multiple of 2 or a multiple of 3.
∴ E_{2}' = {(1,1), (1,2), (2, 1), (1, 3), (2, 2), (3, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6),
(4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (6, 6)}
i.e. n(E_{2}') = 24
Thus, P(E_{2}') = $\frac{24}{36}=\frac{2}{3}$
Hence, required probability P(E_{2}) = 1$-$ P(E_{2}')
$=1-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}$
Page No 33.47:
Question 30:
A bag contains 8 red, 3 white and 9 blue balls. If three balls are drawn at random, determine the probability that (i) all the three balls are blue balls (ii) all the balls are of different colours.
Answer:
Out of 20 balls, three balls can be drawn in ^{20}C_{3} ways.
∴ Total number of elementary events = ^{20}C_{3}
(i)
Out of nine blue balls, three blue balls can be chosen in ^{9}C_{3} ways.
∴ Favourable number of events = ^{9}C_{3} ways.
Hence, required probability = $\frac{{}^{9}C_{3}}{{}^{20}C_{3}}=\frac{9\times 8\times 7}{20\times 19\times 18}=\frac{7}{95}$
(ii)
Out of eight red balls, one red ball can be drawn in ^{8}C_{1} ways.
Out of three white balls, one white ball can be drawn in ^{3}C_{1}.
Out of nine blue balls, one blue ball can be drawn in ^{9}C_{1} ways.
So, favourable number of elementary events = ^{8}C_{1} × ^{3}C_{1} × ^{9}C_{1}
Hence, required probability = $\frac{{}^{8}C_{1}\times {}^{3}C_{1}\times {}^{9}C_{1}}{{}^{20}C_{3}}=\frac{8\times 3\times 9}{60\times 19}=\frac{18}{95}$
Page No 33.47:
Question 31:
A bag contains 5 red, 6 white and 7 black balls. Two balls are drawn at random. What is the probability that both balls are red or both are black?
Answer:
There are 18 balls in the bag out of which two balls can be drawn in ^{18}C_{2} ways.
So, total number of elementary events = ^{18}C_{2} = 153
According to the question, both the balls drwan are either red or black, which means that the two balls should be of the same colour.
Out of five red balls, two red balls can be drawn in ^{5}C_{2} ways.
Similarly, two black balls can be drawn from seven black balls in ^{7}C_{2} ways.
So, favourable number of elementary events = ^{5}C_{2} + ^{7}C_{2} = 10 + 21 = 31
Hence, required probability = $\frac{31}{153}$
Page No 33.47:
Question 32:
If a letter is chosen at random from the English alphabet, find the probability that the letter is (i) a vowel (ii) a consonant
Answer:
We know that there are 26 letters in the English alphabet.
So, total number of elementary events, n(S) = 26
(i)
Out of the five vowels of the English alphabet (a, e, i, o, u), one vowel can be chosen in ^{5}C_{1} ways.
So, favourable number of events = ^{5}C_{1}_{ }= 5
Hence, required probability = $\frac{5}{26}$
(ii)
Out of the 21 consonants of the English alphabet, one consonant can be chosen in ^{21}C_{1} ways.
So, favourable number of events = ^{21}C_{1} = 21
Hence, required probability = $\frac{21}{26}$
Page No 33.47:
Question 33:
In a lottery, a person chooses six different numbers at random from 1 to 20, and if these six numbers match with six number already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?
Answer:
Total number of ways in which one can choose six different numbers from 1 to 20 = ^{20}C_{6}
∴ Total number of elementary events = ^{20}C_{6} = 38760
Hence, there are 38760 combinations of 6 numbers.
Out of these combinations, one is already fixed by the lottery committee.
Favourable number of elementary events = 1
∴ Required probability = $\frac{1}{38760}$
Page No 33.47:
Question 34:
20 cards are numbered from 1 to 20. One card is drawn at random. What is the probability that the number on the cards is:
(i) a multiple of 4?
(ii) not a multiple of 4?
(iii) odd?
(iv) greater than 12?
(v) divisible by 5?
(vi) not a multiple of 6?
Answer:
Clearly, the sample space is given by S = {1, 2, 3, 4, 5........19, 20}.
i.e. n(S) = 20
(i)
Let E_{1} = event of getting a multiple of 4
Then E_{1} = {4, 8, 12, 16, 20}
i.e. n(E_{1}) = 5
Hence, required probability = P(E_{1}) = $\frac{n\left({E}_{1}\right)}{n\left(S\right)}=\frac{5}{20}=\frac{1}{4}$
(ii)
Let E_{2} = event of getting a non-multiple of 4
Then P(non-multiple of 4) = P(E_{2}) = 1 $-$ P(multiple of 4)
= $1-\frac{1}{4}=\frac{3}{4}$
(iii)
Let E_{3} = event of getting an odd number
Then E_{3} = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
i.e. n(E_{3}) = 10
Hence, required probability = P(E_{3}) = $\frac{10}{20}=\frac{1}{2}$
(iv)
Let E_{4}_{ }= event of getting a number greater than 12
Then E_{4} = {13, 14, 15, 16, 17, 18, 19, 20}
i.e. n(E_{4}) = 8
Hence, required probability = P(E_{4}) = $\frac{8}{20}=\frac{2}{5}$
(v)
Let E_{5}_{ }= event of getting a number divisible by 5
Then E_{5} = {5, 10, 15, 20}
i.e. n(E_{5}) = 4
Hence, required probability = P(E_{5}) = $\frac{4}{20}=\frac{1}{5}$
(vi)
Let E_{6}_{ }= event of getting a number which is not a multiple of 6
Then E_{6}' = event of getting a number which is a multiple of 6
E_{6}' = {6, 12, 18}
i.e. n(E_{6}' ) = 3
Now, P(E_{6}') = $\frac{3}{20}$
Hence, required probability P(E_{6}) = 1 − P(E_{6}')
= $1-\frac{3}{20}=\frac{20-3}{20}=\frac{17}{20}$
Page No 33.47:
Question 35:
Two dice are thrown. Find the odds in favour of getting the sum (i) 4 (ii) 5.
(iii) What are the odds against getting the sum 6?
Answer:
(i)
Let A be the event of 'getting the sum 4'.
Then A= {(1, 3), (3, 1), (2, 2)}
Here, there are three favourable outcomes, while there are (36 – 3 =) 33 unfavourable outcomes.
∴ Odds in favour of the sum 4 = $\frac{3}{33}=\frac{1}{11}=1:11$
(ii)
Let A be the event of 'getting the sum 5'.
Then A= {(1, 4), (4, 1), (2, 3), (3, 2)}
Here, there are four favourable outcomes, while there are (36 – 4 =) 32 unfavourable outcomes.
∴ Odds in favour of the sum 5 = $\frac{4}{32}=\frac{1}{8}=1:8$
(iii)
Let A be the event of 'getting the sum 6.
Then A= {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)}
Here, there are five favourable outcomes, while there (36 – 5 =) 31 unfavourable outcomes.
∴ Odds against getting the sum 6 = $\frac{31}{5}=31:5$
Page No 33.47:
Question 36:
What are the odds in favour of getting a spade if the card drawn from a well-shuffled deck of cards? What are the odds in favour of getting a king?
Answer:
In a pack of 52 cards, there are 13 cards of spade.
There are 13 outcomes favourable to the event 'a spade', while the other (52 – 13 =) 39 are unfavourable.
∴ Odds in favour of getting a spade = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{unfavourable}\mathrm{outcomes}}=\frac{13}{39}=\frac{1}{3}=1:3$
Again, there are four kings in a pack of 52 cards.
There are 4 outcomes favourable to the event 'a king', while the other (52 – 4 =) 48 are unfavourable.
∴ Odds in favour of getting a king = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{unfavourable}\mathrm{outcomes}}=\frac{4}{48}=\frac{1}{12}=1:12$
Page No 33.47:
Question 37:
A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn at random. From the box, what is the probability that:
(i) all are blue?
(ii) at least one is green?
Answer:
Out of 60 marbles, five marbles can be drawn in ^{60}C_{5} ways.
∴ Total number of elementary events = ^{60}C_{5}
(i)
Out of 20 blue marbles, five blue marbles can be chosen in ^{20}C_{5} ways.
∴ Favourable number of events = ^{20}C_{5} ways
Hence, the required probability is given by
$\frac{{}^{20}C_{5}}{{}^{60}C_{5}}=\frac{20\times 19\times 18\times 17\times 16}{60\times 59\times 58\times 57\times 56}\phantom{\rule{0ex}{0ex}}=\frac{19\times 6\times 17}{59\times 29\times 57\times 7}\phantom{\rule{0ex}{0ex}}=\frac{2\times 17}{59\times 29\times 7}\phantom{\rule{0ex}{0ex}}=\frac{34}{11977}$
(ii) P (no green) = $\frac{\mathrm{Favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{outcomes}}$
= $\frac{{}^{30}{C}_{5}}{{}^{60}{C}_{5}}$
Thus, P(at least one green) = 1 – P (no green)
$=1-\frac{{}^{30}C_{5}}{{}^{60}C_{5}}\phantom{\rule{0ex}{0ex}}=1-\frac{117}{4484}\phantom{\rule{0ex}{0ex}}=\frac{4484-117}{4484}\phantom{\rule{0ex}{0ex}}=\frac{4367}{4484}$
Page No 33.47:
Question 38:
A box contains 6 red marbles numbered 1 through 6 and 4 white marbles numbered from 12 through 15. Find the probability that a marble drawn is
(i) white
(ii) white and odd numbered
(iii) even numbered
(iv) red or even numbered.
Answer:
Total number of marbles = (6 + 4) = 10
Let S be the sample space.
Then n(S) = number of ways of selecting one marble out of 10 = ^{10}C_{1} = 10 ways
(i)
Let E_{1} = event of getting a white marble
∴ n(E_{1}) = ^{4}C_{1} = 4
Hence, required probability = $\frac{{}^{4}C_{1}}{{}^{10}C_{1}}=\frac{4}{10}=\frac{2}{5}$
(ii)
Let E_{2} = event of getting a white marble, which is odd numbered.
i.e. E_{2} = {13, 15}
∴ n(E_{2}) = 2
Hence, required probability = $\frac{n\left({E}_{2}\right)}{n\left(S\right)}=\frac{2}{10}=\frac{1}{5}$
(iii)
Let E_{3} = event of getting an even numbered marble
i.e. E_{3} = {2, 4, 6, 12, 14}
∴ n(E_{3}) = 5
Hence, required probability = $\frac{n\left({E}_{3}\right)}{n\left(S\right)}=\frac{5}{10}=\frac{1}{2}$
(iv)
Let E_{4} = event of getting a red marble
i.e. E_{4} = {1, 2, 3, 4, 5, 6}
∴ n(E_{4}) = 6
Now, P(E_{4}) = $\frac{6}{10}=\frac{3}{5}$ ...(i)
Let E_{5} = event of getting even numbered marble
Then E_{5}_{ }= {2, 4, 6, 12, 14}
i.e.n(E_{5}) = 5
Now, P(E_{5}) = $\frac{5}{10}=\frac{1}{2}$ ...(ii)
From (i) and (ii), we get:
E_{4} ∩ E_{5} = {2, 4, 6}
$\Rightarrow $n(E_{4} ∩ E_{5}) = 3
$\Rightarrow $P(E_{4} ∩ E_{5}) = $\frac{3}{10}$
By addition theorem, we have:
P (E_{4} ∪ E_{5}) = P(E_{4}) + P (E_{5}) − P (E_{4} ∩ E_{5})
⇒ P (E_{4} ∪ E_{5}) = $\frac{3}{5}+\frac{1}{2}-\frac{3}{10}=\frac{8}{10}=\frac{4}{5}$
Hence, required probability = P(E_{4} ∪ E_{5}) = $\frac{4}{5}$
Page No 33.47:
Question 39:
A class consists of 10 boys and 8 girls. Three students are selected at random. What is the probability that the selected group has
(i) all boys?
(ii) all girls?
(iii) 1 boys and 2 girls?
(iv) at least one girl?
(v) at most one girl?
Answer:
Total number of students = (10 + 8) = 18
Let S be the sample space.
Then n(S) = number of ways of selecting 3 students out of 18 = ^{18}C_{3} ways
(i)
Out of 10 boys, three boys can be selected in ^{10}C_{3} ways.
∴ Favourable number of events, n(E) = ^{10}C_{3}
Hence, required probability = $\frac{{}^{10}C_{3}}{{}^{18}C_{3}}=\frac{10\times 9\times 8}{18\times 17\times 16}=\frac{5}{34}$
(ii)
Out of eight girls, three girls can be selected in ^{8}C_{3} ways.
∴ Favourable number of events, n(E) = ^{8}C_{3}
Hence, required probability = $\frac{{}^{8}C_{3}}{{}^{18}C_{3}}=\frac{8\times 7\times 6}{18\times 17\times 16}=\frac{7}{102}$
(iii)
One boy and two girls can be selected in ^{10}C_{1} × ^{8}C_{2}.
∴ Favourable number of events =^{ }^{10}C_{1} × ^{8}C_{2}
Hence, required probability = $\frac{{}^{10}C_{1}\times {}^{8}C_{2}}{{}^{18}C_{3}}=\frac{10\times 28}{816}=\frac{35}{102}$
(iv)
Probability of at least one girl = 1 $-$ P(no girl)
= 1 $-$ P(all 3 are boys)
= $1-\frac{{}^{10}C_{3}}{{}^{18}C_{3}}=1-\frac{5}{34}=\frac{29}{34}$
(v)
Let E be the event with at most one girl in the group.
Then E = {0 girl, 1 girl}
∴ Favourable number of events, n(E) = ^{8}C_{0} × ^{10}C_{3} × ^{8}C_{1} × ^{10}C_{2}
Hence, the required probability is given by
$\frac{{}^{8}C_{0}\times {}^{10}C_{3}+{}^{8}C_{1}\times {}^{10}C_{2}}{{}^{18}C_{3}}\phantom{\rule{0ex}{0ex}}=\frac{1\times {}^{10}C_{3}+{}^{8}C_{1}\times {}^{10}C_{2}}{{}^{18}C_{3}}\phantom{\rule{0ex}{0ex}}=\frac{1\times 120+8\times 45}{816}\phantom{\rule{0ex}{0ex}}=\frac{480}{816}\phantom{\rule{0ex}{0ex}}=\frac{10}{17}$
Page No 33.48:
Question 40:
Five cards are drawn from a well-shuffled pack of 52 cards. Find the probability that all the five cards are hearts.
Answer:
Out of 52 cards from a deck, 5 cards can be drawn in ^{52}C_{5} ways.
∴ Total number of elementary events = ^{52}C_{5}
Out of 13 cards of heart, 5 cards can be drawn in ^{13}C_{5} ways.
∴ Favourable number of events = ^{13}C_{5}
Hence, required probability = $\frac{{}^{13}C_{5}}{{}^{52}C_{5}}=\frac{13\times 12\times 11\times 10\times 9}{52\times 51\times 50\times 49\times 48}=\frac{33}{66640}$
Page No 33.48:
Question 41:
A bag contains tickets numbered from 1 to 20. Two tickets are drawn. Find the probability that (i) both the tickets have prime numbers on them (ii) on one there is a prime number and on the other there is a multiple of 4.
Answer:
Clearly, the sample space is given by S = {1, 2, 3, 4, 5,...19, 20}.
∴ n(S) = ^{20}C_{2} = 190
(i)
Let E_{1} be the event where both the tickets have prime numbers on them.
Then E_{1} = {2, 3, 5, 7, 11, 13, 17, 19}
∴ Favourable number of ways = n(E_{1}) = ^{8}C_{2}
Hence, required probability = P(E_{1}) = $\frac{n\left({E}_{1}\right)}{n\left(S\right)}=\frac{{}^{8}C_{2}}{{}^{20}C_{2}}=\frac{14}{95}$
(ii)
Let E_{2} be the event where one ticket has a prime number, while the other has a multiple of 4.
Then prime numbers = {2, 3, 5, 7, 11, 13, 17, 19}
and multiples of 4 = {4, 8, 12, 16, 20}
∴ Favourable number of ways, n(E_{2}) = ^{8}C_{1}× ^{5}C_{1} = 8 × 5 = 40
Hence, required probability, P(E_{2}_{ }) = _{$\frac{n\left({E}_{2}\right)}{n\left(S\right)}=\frac{40}{190}=\frac{4}{19}$}
Page No 33.48:
Question 42:
An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that
(i) both the balls are red
(ii) one ball is red and the other is black
(iii) one ball is white.
Answer:
Out of 15 balls, two balls can be drawn in ^{15}C_{2} ways.
∴ Total number of elementary events = ^{15}C_{2} = 105
(i)
Out of three red balls, two red balls can be chosen in ^{3}C_{2} ways.
∴ Favourable number of ways = ^{3}C_{2} = 3
Hence, required probability =
(ii)
Out of three red balls, one red ball can be drawn in ^{3}C_{1} ways; and out of five black balls, one black ball can be drawn in ^{5}C_{1} ways.
Therefore, one red and one black can be drawn in ^{3}C_{1}× ^{5}C_{1} ways.
∴ Favourable number of ways = ^{3}C_{1}× ^{5}C_{1} = 3× 5 = 15
Hence, required probability = $\frac{15}{105}=\frac{1}{7}$
(iii)
Out of seven white balls, one white ball can be drawn in ^{7}C_{1} ways; and one ball can be drawn from the rest of the other coloured (red and black) balls in ^{8}C_{1} ways.
∴ Favourable number of ways = ^{7}C_{1}× ^{8}C_{1} = 7× 8 = 56
Hence, required probability = $\frac{56}{105}=\frac{8}{15}$
Page No 33.48:
Question 43:
A and B throw a pair of dice. If A throws 9, find B's chance of throwing a higher number.
Answer:
A and B throw a pair of dices.
Then all the possible outcomes = 6^{2} = 36
i.e. total number of possible outcome, n(S) = 36
Consider E = event where A throws 9 and B throws more than 9, i.e. 10, 11 and 12
E = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
i.e. n(E) = 6
Hence, required probability = $\frac{6}{36}=\frac{1}{6}$
Page No 33.48:
Question 44:
In a hand at Whist, what is the probability that four kings are held by a specified player?
Answer:
In a hand at whist, the specified players have 13 cards including fourkings.
Four kings can be drawn out of four in ^{4}C_{4} ways.
The remaining nine cards held by him can be drawn out of the remaining 48 cards in the pack in ^{48}C_{9} ways.
The total number of ways in which four kings and nine other cards can be drawn is ^{4}C_{4}× ^{48}C_{9}_{. }
Favourable number of cases = ^{4}C_{4}× ^{48}C_{9}_{ }
Also, exhaustive number of cases = ^{52}C_{13}
∴ Required probability = $\frac{{}^{4}C_{4}\times {}^{48}C_{9}}{{}^{52}C_{13}}=\frac{11}{4165}$
Page No 33.48:
Question 45:
Find the probability that in a random arrangement of the letters of the word 'UNIVERSITY', the two I's do not come together.
Answer:
Out of the letters in the word ‘UNIVERSITY’, there are two I’s.
Number of permutations = $\frac{10!}{2}$
The number of words in which two I’s are never together is given by
total number of words – number of words in which two I’s are together.
$=\frac{10!}{2}-9!\phantom{\rule{0ex}{0ex}}=\frac{10!-2\times 9!}{2}\phantom{\rule{0ex}{0ex}}=\frac{9!\left[10-2\right]}{2}\phantom{\rule{0ex}{0ex}}=\frac{9!\times 8}{2}\phantom{\rule{0ex}{0ex}}=9!\times 4$
∴ Required probability = $\frac{9!\times 4}{{\displaystyle \frac{10!}{2}}}=\frac{9!\times 4\times 2}{10\times 9!}=\frac{8}{10}=\frac{4}{5}$
Page No 33.6:
Question 1:
A coin is tossed once. Write its sample space
Answer:
If a coin is tossed once, the possible outcomes are a head (H) and tail (T).
Therefore, the sample space of this experiment is S = { H, T}.
Page No 33.6:
Question 2:
If a coin is tossed two times, describe the sample space associated to this experiment.
Answer:
If a coin is tossed twice, the possible outcomes are HH, HT, TH, TT.
Therefore, the sample space of this experiment is S = {HH, HT, TH, TT}.
Page No 33.6:
Question 3:
If a coin is tossed three times (or three coins are tossed together), then describe the sample space for this experiment.
Answer:
A coin has two faces: a head (H) and a tail (T).
When a coin is tossed three times, the total number of possible outcomes is 2^{3} = 8.
Thus, when a coin is tossed three times, the sample space is given by S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
Page No 33.6:
Question 4:
Write the sample space for the experiment of tossing a coin four times.
Answer:
When a coin is tossed once, there are two possible outcomes: a head (H) and a tail (T).
When a coin is tossed four times, the total number of possible outcomes is 2^{4} = 16.
Thus, when a coin is tossed four times, the sample space is given by
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
Page No 33.6:
Question 5:
Two dice are thrown. Describe the sample space of this experiment.
Answer:
When two dices are thrown, there are (6 × 6) = 36 outcomes.
The set of these outcomes is the sample space, which is given by
S = (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Page No 33.6:
Question 6:
What is the total number of elementary events associated to the random experiment of throwing three dice together?
Answer:
When we throw a dice, it can result in any of the six numbers, namely 1, 2, 3, 4, 5 and 6.
When three dices are thrown together, the total number of elementary events associated is 6^{3} = ( 6 × 6 × 6) = 216.
Page No 33.6:
Question 7:
A coin is tossed and then a die is thrown. Describe the sample space for this experiment.
Answer:
When a coin is tossed, the outcomes are H, T. When a dice is thrown, the outcomes can be 1, 2, 3, 4, 5, 6.
Thus, we can write the two parts of this experiment as follows:
Hence, the sample space is given by
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6),
(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
Page No 33.6:
Question 8:
A coin is tossed and then a die is rolled only in case a head is shown on the coin. Describe the sample space for this experiment.
Answer:
A coin has two faces: a head (H) and a tail (T).
A dice has six faces that are numbered from 1, 2, 3, 4, 5, 6 with one number on each face.
When a head is shown on a coin toss, a dice is rolled.
Thus, the sample space is given by
S = {(H, 1),(H, 2), (H, 3), (H, 4), (H, 5), (H, 6), T}
Page No 33.6:
Question 9:
A coin is tossed twice. If the second throw results in a tail, a die is thrown. Describe the sample space for this experiment.
Answer:
When a coin is tossed twice, the possible outcomes are HH, HT, TH, TT.
The second throw results in a head in HH, TH.
The second throw results in a tail in HT, TT.
Now, a dice is thrown.
The possible outcomes on a dice are 1, 2, 3, 4, 5 and 6.
The sample space is given by
S = { HH, TH, (HT, 1), (HT, 2), (HT, 3), (HT, 4), (HT, 5), (HT, 6), (TT, 1), (TT, 2), (TT, 3) (TT, 4), (TT, 5), (TT 6)}.
∴ The total number of elements of the sample space is 14.
Page No 33.6:
Question 10:
An experiment consists of tossing a coin and then tossing it second time if head occurs. If a tail occurs on the first toss, then a die is tossed once. Find the sample space.
Answer:
A coin has two faces: a head (H) and a tail (T).
A dice has six faces that are numbered from 1 to 6, with one number on each face.
Thus, the sample space of the given experiment is given by
S = {(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
Page No 33.67:
Question 1:
(a) If A and B be mutually exclusive events associated with a random experiment such that P(A) = 0.4 and P(B) = 0.5, then find:
(i) P (A ∪ B)
(ii) $P(\overline{)A}\cap \overline{)B})$
(iii) P ($\overline{)A}$ ∩ B)
(iv) P (A ∩ $\overline{)B}$).
(b) A and B are two events such that P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35.
Find (i) P (A ∪ B)
(ii) $P(\overline{)A}\cap \overline{)B)}$
(iii) P (A ∩ $\overline{)B}$)
(iv) P (B ∩ $\overline{)A}$)
(c) Fill in the blanks in the following table:
P (A) | P (B) | P (A ∩ B) | P (A ∪ B) | |
(i) | $\frac{1}{3}$ | $\frac{1}{5}$ | $\frac{1}{15}$ | ...... |
(ii) | 0.35 | .... | 0.25 | 0.6 |
(iii) | 0.5 | 0.35 | ..... | 0.7 |
Answer:
(a)
Given:
P(A) = 0.4 and P(B) = 0.5
If A and B be mutually exclusive event, then P (A ∩ B) = 0
(i)
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
= 0.4 + 0.5 $-$ 0 = 0.9
(ii)
$P\left(\overline{A}\cap \overline{)B}\right)=1-P\left(A\cup B\right)$
= 1 $-$ 0.9 = 0.1
(iii)
$P\left(\overline{A}\cap B\right)=P\left(B\right)-P\left(A\cap B\right)$
= 0.5 $-$ 0 = 0.5
(iv)
$P\left(A\cap \overline{B}\right)=P\left(A\right)-P\left(A\cap B\right)$
= 0.4 $-$ 0 = 0.4
(b)
Given:
P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35
(i)
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
= 0.54 + 0.69 $-$ 0.35
= 0.88
(ii)
$P\left(\overline{A}\cap \overline{)B}\right)=1-P\left(A\cup B\right)$
= 1 $-$ 0.88
= 0.12
(iii)
$P\left(A\cap \overline{B}\right)=P\left(A\right)-P\left(A\cap B\right)$
= 0.54 $-$ 0.35
= 0.19
(iv)
$P\left(\overline{A}\cap B\right)=P\left(B\right)-P\left(A\cap B\right)$
= 0.69 $-$ 0.35
= 0.34
(c)
Given:
$P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{5}\mathrm{and}\mathit{}P\left(\mathit{A}\mathit{\cap}\mathit{B}\right)=\frac{1}{15}$
(i)
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
$=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}\phantom{\rule{0ex}{0ex}}$
$=\frac{5+3-1}{15}=\frac{7}{15}$
(ii)
Given:
P (A) = 0.35, P (A ∪ B) = 0.6 and P (A ∩ B) = 0.25
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B)$-$ P (A ∩ B)
0.6 = 0.35 + P (B) $-$ 0.25
P (B) = 0.6 $-$ 0.35 + 0.25
= 0.6 $-$ 0.1 = 0.5
(iii)
Given:
P (A) = 0.5, P(B) = 0.35 and P (A ∪ B) = 0.7
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
0.7 = 0.5 + 0.35 $-$ P (A ∩ B)
P (A ∩ B) = 0.5 + 0.35 $-$ 0.7
= 0.85 $-$ 0.7 = 0.15
Page No 33.68:
Question 2:
If A and B are two events associated with a random experiment such that
P(A) = 0.3, P (B) = 0.4 and P (A ∪ B) = 0.5, find P (A ∩ B).
Answer:
Given:
P(A) = 0.3, P (B) = 0.4 and P (A ∪ B) = 0.5
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
⇒ 0.5 = 0.3 + 0.4 $-$ P (A ∩ B)
⇒ 0.5 = 0.7 $-$ P (A ∩ B)
⇒ P (A ∩ B) = 0.7 $-$ 0.5
= 0.2
Hence, P (A ∩ B) = 0.2
Page No 33.68:
Question 3:
If A and B are two events associated with a random experiment such that
P(A) = 0.5, P(B) = 0.3 and P (A ∩ B) = 0.2, find P (A ∪ B).
Answer:
Given:
P(A) = 0.5, P (B) = 0.3 and P (A ∩ B) = 0.2
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
⇒ P (A ∪ B) = 0.5 + 0.3 $-$ 0.2
= 0.8 $-$ 0.2 = 0.6
Hence, P (A ∪ B) = 0.6
Page No 33.68:
Question 4:
If A and B are two events associated with a random experiment such that
P (A ∪ B) = 0.8, P (A ∩ B) = 0.3 and P $\overline{)\left(A\right)}$ = 0.5, find P(B).
Answer:
Given:
P (A ∪ B) = 0.8, P (A ∩ B) = 0.3 and $P\left(\overline{A}\right)=0.5$
We know that
$P\left(A\right)+P\left(\overline{A}\right)=1$
$\Rightarrow P\left(A\right)+0.5=1\phantom{\rule{0ex}{0ex}}$
$\Rightarrow P\left(A\right)=1-0.5=0.5$
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$P (A ∩ B)
⇒ 0.8 = 0.5 + P (B) $-$ 0.3
⇒ 0.8 = 0.2 + P (B)
⇒ P (B) = 0.8 $-$ 0.2
= 0.6
Hence, P (B) = 0.6
Page No 33.68:
Question 5:
Given two mutually exclusive events A and B such that P(A) = 1/2 and P(B) = 1/3, find P (A or B).
Answer:
Given:
P(A) = 1/2 and P(B) = 1/3
For mutually exclusive events A and B, we have:
P(A or B) = P(A) + P(B)
$=\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}=\frac{5}{6}$
Hence, $P\left(AorB\right)=\frac{5}{6}$
Page No 33.68:
Question 6:
There are three events A, B, C one of which must and only one can happen, the odds are 8 to 3 against A, 5 to 2 against B, find the odds against C.
Answer:
Since the odds against event A are 8 : 3, the probability of the happening of event A is given by
P(A) = $\frac{3}{8+3}=\frac{3}{11}$
Similarly, the odds against event B are 5 : 2.
So, P(B) = $\frac{2}{5+2}=\frac{2}{7}$
Since events A, B and C are such that one of them must and only one can happen, A, B and C are mutually exclusive and totally exhaustive events.
Consequently, A ∪ B ∪ C = S
and A ∩ B = B ∩ C = C ∩ A = Φ
Thus, P (A ∪ B ∪ C) = P(S) = 1
⇒ P(A) + P(B) + P(C) = 1
⇒ $\frac{3}{11}+\frac{2}{7}+P\left(C\right)=1\phantom{\rule{0ex}{0ex}}$
$\Rightarrow P\left(C\right)=1-\frac{3}{11}-\frac{2}{7}=\frac{34}{77}$
Hence, odds against the events are $P\left(\overline{C}\right):P\left(C\right)=\left(1-\frac{34}{77}\right):\frac{34}{77}$
= (77 $-$ 34) : 34
= 43 : 34
Page No 33.68:
Question 7:
One of the two events must happen. Given that the chance of one is two-third of the other, find the odds in favour of the other.
Answer:
Let the given events be A and B
Now, $P\left(A\right)=\frac{2}{3}P\left(B\right)$.
Let P(B) = x
$\therefore P\left(A\right)=\frac{2}{3}x$
The events A and B are exhaustive.
∴ P(A) or P(B) = 1
⇒ P(A) + P(B) = 1 (∵ A and B are mutually exclusive)
$\Rightarrow \frac{2}{3}x+x=1\phantom{\rule{0ex}{0ex}}$
$\Rightarrow \frac{5x}{3}=1\phantom{\rule{0ex}{0ex}}$
$\Rightarrow x=\frac{3}{5}\phantom{\rule{0ex}{0ex}}$
$\therefore P\left(B\right)=\frac{3}{5}\phantom{\rule{0ex}{0ex}}$
This implies that the odds in favour of B are 3 : (5 – 2), i.e. 3 : 2.
Page No 33.68:
Question 8:
A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of its being a spade or a king.
Answer:
If A and B denote the events of drawing a spade card and a king, respectively, then event A consists of 13 sample points, whereas event B consists of four sample points.
Thus, $P\left(A\right)=\frac{13}{52}$ and $P\left(B\right)=\frac{4}{52}$
The compound event (A ∩ B) consists of only one sample point, i.e. the king of spade.
So, $P\left(A\cap B\right)=\frac{1}{52}$
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B)$-$ P (A ∩ B)
= $\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{13+4-1}{52}=\frac{16}{52}=\frac{4}{13}$
Hence, the probability that the card drawn is either a spade or a king is given by $\frac{4}{13}.$
Page No 33.68:
Question 9:
In a single throw of two dice, find the probability that neither a doublet nor a total of 9 will appear.
Answer:
Here, S = {(1, 1), (1, 2)...,(6, 5), (6, 6)}
∴ Number of possible outcomes = 6 × 6 = 36
Let E be the event where a doublet appears and F be the event where the total is 9.
i.e. E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
and F = {(3, 6), (4, 5), (5, 4), (6, 3)}
$\therefore P\left(E\right)=\frac{6}{36}=\frac{1}{6}\mathrm{and}P\left(\mathrm{F}\right)=\frac{4}{36}=\frac{1}{9}$
P(neither a doublet nor a total of 9) = $P\left(\overline{E}\cap \overline{F}\right)=P\left(\overline{\left(E\cup F\right)}\right)=1-P\left(E\cup F\right)$ ...(i)
The events E and F are mutually exclusive.
By addition theorem, we have:
P (E ∪ F) = P(E) + P (F)
= $\frac{1}{6}+\frac{1}{9}=\frac{3+2}{18}=\frac{5}{18}$
From (i), we get:
$P\left(\overline{E}\cap \overline{F}\right)=1-\frac{5}{18}=\frac{18-5}{18}=\frac{13}{18}$
Page No 33.68:
Question 10:
A natural number is chosen at random from amongst first 500. What is the probability that the number so chosen is divisible by 3 or 5?
Answer:
Let S be the sample space.
Then n(S) = 500
∴ Total number of elementary events = 500
Let A be the event where the number selected is divisible by 3 and B be the event where the number selected is divisible by 5.
Then A = {3, 6, 9, 12, 15, ...498}
and B = {5, 10, 15, 20, 25, ...500}
and (A ∩ B) = { 15, 30, 45, ...495}
We have:
$n\left(A\right)=\frac{498}{3}=166$
$n\left(B\right)=\frac{500}{5}=100\phantom{\rule{0ex}{0ex}}$
$n\left(A\cap B\right)=\frac{495}{15}=33$ [∵ LCM of 3 and 5 is 15]
$\therefore P\left(A\right)=\frac{166}{500},P\left(B\right)=\frac{100}{500}\mathrm{and}\mathit{}P\left(A\mathit{\cap}B\right)=\frac{33}{500}$
Now, required probability = P(a number is divisible by 3 or 5)
= P (A ∪ B)
= P(A) + P(B) $-$ P(A ∩ B)
= $\frac{166}{500}+\frac{100}{500}-\frac{33}{500}=\frac{266-33}{500}=\frac{233}{500}$
Page No 33.68:
Question 11:
A dice is thrown twice. What is the probability that at least one of the two throws come up with the number 3?
Answer:
Let A be event where the first throw comes up with the number 3.
Also, let B be the event where the second throw comes up with the number 3.
∴ Favourable events of A = {(3, 1),(3, 2), (3, 3), (3, 4), (3, 5), (3, 6)} = 6
$\Rightarrow P\left(A\right)=\frac{6}{36}=\frac{1}{6}$
∴ Favourable events of B = {(1, 3), (2, 3), (3, 3), (4, 3),(5, 3), (6, 3) } = 6
$\Rightarrow P\left(B\right)=\frac{6}{36}=\frac{1}{6}$
Also, $P\left(A\cap B\right)=\frac{1}{36}$ [∵ Only one event will be common, i.e. (3, 3)]
Hence, required probability =$P(A\cup B)$
$=P\left(A\right)+P\left(B\right)-P(A\cap B)\phantom{\rule{0ex}{0ex}}=\frac{1}{6}+\frac{1}{6}-\frac{1}{36}\phantom{\rule{0ex}{0ex}}=\frac{6+6-1}{36}\phantom{\rule{0ex}{0ex}}=\frac{11}{36}$
Page No 33.68:
Question 12:
A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.
Answer:
If A and B denote the events of drawing an ace and a spade card, respectively, then event A consists of four sample points, whereas event B consists of 13 sample points.
Thus, $P\left(A\right)=\frac{4}{52}$ and $P\left(B\right)=\frac{13}{52}$
The compound event (A ∩ B) consists of only one sample point, i.e. an ace of spade.
So, $P\left(A\cap B\right)=\frac{1}{52}$
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
= $\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{4+13-1}{52}=\frac{16}{52}=\frac{4}{13}$
Hence, the probability that the card drawn is either an ace or a spade card is given by $\frac{4}{13}.$
Page No 33.68:
Question 13:
The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English Examination is 0.75. What is the probability of passing the Hindi Examination?
Answer:
Let A and B be the events of passing English and Hindi examinations, respectively.
Accordingly, we have:
P(A and B) = 0.5
P(not A and not B) = 0.1 [i.e. P(A' ∩ B') = 0.1]
P(A) = 0.75
Now, P(A∪B) + P(A' ∩ B') = 1
⇒ P(A∪B) = 1$-$ P(A' ∩ B')
= 1 $-$ 0.1 = 0.9
By addition theorem, we have:
P(A ∪ B) = P(A) + P(B) $-$ P(A ∩ B)
⇒ 0.9 = 0.75 + P (B) $-$ 0.5
⇒ P(B) = 0.9 $-$ 0.75 + 0.5
⇒ P(B) = 0.65
Thus, the probability of passing the Hindi examination is 0.65.
Page No 33.68:
Question 14:
One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?
Answer:
Let S be the sample space.
Then n(S) = 100
∴ Total number of elementary events = 100
Let A be the event where the number selected is divisible by 4 and B be the event where the number selected is divisible by 6.
Then A = {4, 8, 12, 16, ...100 }
B = { 6, 12, 18, 24, ...96},
and (A ∩ B) = {12, 24, ...96}
Now, we have:
$n\left(A\right)=\frac{100}{4}=25$
$n\left(B\right)=\frac{96}{6}=16$
$n\left(A\cap B\right)=\frac{96}{12}=8$ [∵ LCM of 4 and 6 is 12]
$\therefore P\left(A\right)=\frac{25}{100},P\left(B\right)=\frac{16}{100}\mathrm{and}P\left(A\cap B\right)=\frac{8}{100}$
Now, required probability = P(a number is divisible by 4 or 6)
= P (A ∪ B)
= P(A) + P(B) $-$P(A ∩ B)
= $\frac{25}{100}+\frac{16}{100}-\frac{8}{100}=\frac{25+16-8}{100}=\frac{33}{100}$
Page No 33.68:
Question 15:
From a well shuffled deck of 52 cards, 4 cards are drawn at random. What is the probability that all the drawn cards are of the same colour.
Answer:
Out of 52 cards, four cards can be randomly chosen in ^{52}C_{4} ways.
∴ n(S) = ^{52}C_{4}
Let A = event where the four cards drawn are red
and B = event where the four cards drawn are black
Then, n(A) = ^{26}C_{4} and n(B) = ^{26}C_{4}
$\Rightarrow P\left(A\right)=\frac{{}^{26}C_{4}}{{}^{52}C_{4}}$ and $P\left(B\right)=\frac{{}^{26}C_{4}}{{}^{52}C_{4}}$
A and B are mutually exclusive events.
i.e. P (A ∩ B) = 0
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
= $\frac{{}^{26}C_{4}}{{}^{52}C_{4}}+\frac{{}^{26}C_{4}}{{}^{52}C_{4}}$$-$ 0
= $\frac{46}{17\times 49}+\frac{46}{17\times 49}\phantom{\rule{0ex}{0ex}}$
= $2\times \frac{46}{17\times 49}=\frac{92}{833}$
Hence, the probability that all the drawn cards are of the same colour is $\frac{92}{833}$.
Page No 33.68:
Question 16:
100 students appeared for two examination, 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has passed at least one examination.
Answer:
Let S be the sample space associated with the experiment of students who appeared for two examination.
Then n(S) = 100
∴ Total number of elementary events = 100
Consider the following events:
A = students passed in first examination
B = students passed in second examination
Then n(A) = 60 and n(B) = 50 and n(A ∩ B) = 30
$\therefore P\left(A\right)=\frac{60}{100},P\left(B\right)=\frac{50}{100}$ and $P\left(A\cap B\right)=\frac{30}{100}$
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
= $\frac{50}{100}+\frac{60}{100}-\frac{30}{100}=\frac{50+60-30}{100}=\frac{80}{100}=\frac{4}{5}$
Hence, the probability that a student selected at random has passed at least one examination is $\frac{4}{5}.$
Page No 33.68:
Question 17:
A box contains 10 white, 6 red and 10 black balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either white or red?
Answer:
There are 10 + 6 + 10 = 26 balls in total.
So, the total number of possible outcomes is 26.
Consider the following events:
W = event of drawing a white ball
R = event of drawing a red ball
Then n(W) = 10 and n(R) = 6
Since both the events are mutually exclusive, we have:
(A ∩ B) = 0
$\therefore P\left(A\right)=\frac{10}{26},P\left(B\right)=\frac{6}{26}$ and P (A ∩ B) = 0
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$P (A ∩ B)
= $\frac{10}{26}+\frac{6}{26}-0\phantom{\rule{0ex}{0ex}}$
= $\frac{16}{26}=\frac{8}{13}$
Hence, the probability that the ball drawn is either white or red is $\frac{8}{13}$.
Page No 33.68:
Question 18:
In a race, the odds in favour of horses A, B, C, D are 1 : 3, 1 : 4, 1 : 5 and 1 : 6 respectively. Find probability that one of them wins the race.
Answer:
Let Q, R, S and T be the events where horses A, B, C and D win the race, respectively.
Then, $P\left(Q\right)=\frac{1}{4},P\left(R\right)=\frac{1}{5},P\left(S\right)=\frac{1}{6}\mathrm{and}P\left(T\right)=\frac{1}{7}$
Since only one horse can win the race, Q, R, S and T are mutually exclusive events.
∴ Required probability = P (Q ∪ R ∪ S ∪ T)
= P(Q) + P(R) + P(S) + P(T)
= $\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}$
= $\frac{319}{420}$
Page No 33.68:
Question 19:
The probability that a person will travel by plane is 3/5 and that he will travel by trains is 1/4. What is the probability that he (she) will travel by plane or train?
Answer:
We have two events such that
A = a person will travel by plane
and B = a person will travel by train.
i.e. $P\left(A\right)=\frac{3}{5}$ and $P\left(B\right)=\frac{1}{4}$
Since A and B are mutually exclusive events, we have:
P (A ∩ B) = 0
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
= $\frac{3}{5}+\frac{1}{4}-0=\frac{12+5}{20}=\frac{17}{20}$
Hence, required probability = $\frac{17}{20}$
Page No 33.68:
Question 20:
Two cards are drawn from a well shuffled pack of 52 cards. Find the probability that either both are black or both are kings.
Answer:
Let S be the sample space.
Then n(S) = number of ways of drawing two cards out of 52 =^{ }^{52}C_{2}
Let E_{1} = event in which both the cards are black cards
and E_{2} = event in which both the cards are kings.
Then (E_{1}∩ E_{2}) = event of getting two black kings
i.e. n(E_{1}) = number of ways of drawing two black cards out of the black cards = ^{26}C_{2}
n(E_{2}) = number of ways of drawing two kings out of four kings = ^{4}C_{2}
∴ n(E_{1}∩ E_{2}) = number of ways of drawing two black kings out of two kings = ^{2}C_{2} = 1
Thus, $P\left({E}_{1}\right)=\frac{n\left({E}_{1}\right)}{n\left(S\right)}=\frac{{}^{26}C_{2}}{{}^{52}C_{2}};P\left({E}_{2}\right)=\frac{n\left({E}_{2}\right)}{n\left(S\right)}=\frac{{}^{4}C_{2}}{{}^{52}C_{2}}\phantom{\rule{0ex}{0ex}}$
and $P\left({E}_{1}\cap {E}_{2}\right)=\frac{n\left({E}_{1}\cap {E}_{2}\right)}{n\left(S\right)}=\frac{1}{{}^{52}C_{2}}$
∴ P(drawing both red cards or both kings) = P(E_{1}or E_{2})
= P(E_{1}∪ E_{2})
= P(E_{1}) + P(E_{2})$-$ P( E_{1}∩ E_{2})
= $\left(\frac{{}^{26}C_{2}}{{}^{52}C_{2}}+\frac{{}^{4}C_{2}}{{}^{52}C_{2}}-\frac{1}{{}^{52}C_{2}}\right)\phantom{\rule{0ex}{0ex}}$
= $\frac{\left({}^{26}C_{2}+{}^{4}C_{2}-1\right)}{{}^{52}C_{2}}$
= $\frac{\left(326+6+1\right)}{1326}=\frac{330}{1326}=\frac{55}{221}$
Hence, the required probability is $\frac{55}{221}$.
Page No 33.69:
Question 21:
In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?
Answer:
Let A and B be the events of passing the first and the second examinations, respectively.
Accordingly, P(A) = 0.8, P(B) = 0.7 and P(A or B) = 0.95
We know that
P(A or B) = P(A) + P(B) – P(A and B)
⇒ 0.95 = 0.8 + 0.7 – P(A and B)
⇒ P(A and B) = 0.8 + 0.7 – 0.95
= 1.5 - 0.95 = 0.55
Thus, the probability of passing both the examinations is 0.55.
Page No 33.69:
Question 22:
A box contains 30 bolts and 40 nuts. Half of the bolts and half of the nuts are rusted. If two items are drawn at random, what is the probability that either both are rusted or both are bolts?
Answer:
The numbers of bolts and nuts are 30 and 40, respectively.
Then the numbers of rusted bolts and rusted nuts are 15 and 20, respectively.
Total number of items = 30 + 40 = 70
Total number of rusted items = 15 + 20 = 35
Total number of ways of drawing two items = ^{70}C_{2}
Let R and B be the events in which both the items drawn are rusted items and bolts, respectively.
R and B are not mutually exclusive events, because there are 15 rusted bolts.
P(items are both rusted or both bolts) = P (R ∪ B)
= P(R) + P (B) $-$P (R ∩ B)
= $\frac{{}^{35}C_{2}}{{}^{70}C_{2}}+\frac{{}^{30}C_{2}}{{}^{70}C_{2}}-\frac{{}^{15}C_{2}}{{}^{70}C_{2}}=\frac{1\times 2}{70\times 69}\left(\frac{35\times 34}{1\times 2}+\frac{30\times 29}{1\times 2}-\frac{15\times 14}{1\times 2}\right)\phantom{\rule{0ex}{0ex}}$
$=\frac{1850}{70\times 69}=\frac{185}{483}$
Page No 33.69:
Question 23:
An integer is chosen at random from first 200 positive integers. Find the probability that the integer is divisible by 6 or 8.
Answer:
Let S be the sample space. Then n(S) = 200
∴ Total number of elementary events = 200
Let A be the event in which the number selected is divisible by 6 and B be the event in which the number selected is divisible by 8.
Then A = {6, 12, 18, 24, ...198 },
B = { 8, 16, 24, 32, ...200}
and (A ∩ B) = {24, 48, 72, ...192}
Now, we have:
$n\left(A\right)=\frac{198}{6}=33$
$n\left(B\right)=\frac{200}{8}=25$
$n\left(A\cap B\right)=\frac{192}{24}=8$ [∵ LCM of 6 and 8 is 24]
$\therefore P\left(A\right)=\frac{33}{200},P\left(B\right)=\frac{25}{200}\mathrm{and}P\left(A\cap B\right)=\frac{8}{200}$
Now, required probability = P(a number is divisible by 6 or 8)
= P (A ∪ B)
= P(A) + P(B) $-$ P(A ∩ B)
= $\frac{33}{200}+\frac{25}{200}-\frac{8}{200}=\frac{33+25-8}{200}=\frac{50}{200}=\frac{1}{4}$
Page No 33.69:
Question 24:
Find the probability of getting 2 or 3 tails when a coin is tossed four times.
Answer:
Let S be the sample space associated with the experiment that a coin is tossed four times.
Then n(S) = 2^{4} = 16
Consider the following events:
A: Event of getting 2 tails
B : Event of getting 3 tails
Then A = {HHTT , HTHT, HTTH, THTH, TTHH, THHT}
n(A) = 6
$\therefore P\left(A\right)=\frac{6}{16}$
B = {HTTT, THTT, TTHT, TTTH}
n (B) = 4
$\therefore P\left(B\right)=\frac{4}{16}$
Since events A and B are mutually exclusive, we have:
$P\left(A\cap B\right)=0$
By addition theorem, we have:
P(A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
= $\frac{6}{16}+\frac{4}{16}-0=\frac{10}{16}=\frac{5}{8}$
Page No 33.69:
Question 25:
Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9. [NCERT EXEMPLAR]
Answer:
Let S be the sample space. Then
Total number of elementary events, n(S) = 1000
Let A be the event that the number selected is a multiple of 2 and B be the event that the number selected is a multiple of 9. Then,
A = {2, 4, 6, ..., 998, 1000}
B = {9, 18, 27, ..., 990, 999}
Now, A ∩ B is the event that the number selected is a multiple of 2 and 9 i.e. 18.
A ∩ B = {18, 36, ..., 990}
We have,
n(A) = $\frac{1000}{2}=500$, n(B) = $\frac{999}{9}=111$ and n(A ∩ B) = $\frac{990}{18}=55$
∴ P(A) = $\frac{500}{1000}$, P(B) = $\frac{111}{1000}$ and P(A ∩ B) = $\frac{55}{1000}$
Now,
P(integer is a multiple of 2 or a multiple of 9)
= P(A ∪ B)
= P(A) + P(B) − P(A ∩ B)
= $\frac{500}{1000}+\frac{111}{1000}-\frac{55}{1000}$
= $\frac{556}{1000}$
= 0.556
Hence, the required probability is 0.556.
Page No 33.69:
Question 26:
In a large metropolitan area, the probabilities are 0.87, 0.36, 0.30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either any one or both kinds of sets? [NCERT EXEMPLAR]
Answer:
Let A be the event that a family owns a colour television set and B be the event that a family owns a black and white television set.
So, A ∩ B is the event that a family owns both a colour television set and a black and white television set.
∴ P(A) = 0.87, P(B) = 0.36 and P(A ∩ B) = 0.30 (Given)
Now,
P(a family owns either any one or both kinds of television sets)
= P(A ∪ B)
= P(A) + P(B) − P(A ∩ B)
= 0.87 + 0.36 − 0.30
= 0.93
Hence, the required probability is 0.93.
Page No 33.69:
Question 27:
If A and B are mutually exclusive events such that P(A) = 0.35 and P(B) = 0.45, find
(i) P(A ∪ B) (ii) P(A ∩ B) (iii) P(A ∩ $\overline{)B}$) (iv) P( $\overline{)A}$ ∩ $\overline{)B}$) [NCERT EXEMPLAR]
Answer:
It is given that A and B are mutually exclusive events.
∴ P(A ∩ B) = 0
Also, P(A) = 0.35 and P(B) = 0.45.
(i)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
= 0.35 + 0.45 − 0
= 0.80
(ii) A and B are mutually exclusive events.
∴ P(A ∩ B) = 0
(iii)
$\mathrm{P}\left(A\cap \overline{)B}\right)=\mathrm{P}\left(A\right)-\mathrm{P}\left(A\cap B\right)\phantom{\rule{0ex}{0ex}}=0.35-0\phantom{\rule{0ex}{0ex}}=0.35$
(iv)
$\mathrm{P}\left(\overline{)A}\cap \overline{)B}\right)=\mathrm{P}\left(\overline{)A\cup B}\right)\phantom{\rule{0ex}{0ex}}=1-\mathrm{P}\left(A\cup B\right)\phantom{\rule{0ex}{0ex}}=1-0.80\phantom{\rule{0ex}{0ex}}=0.20$
Page No 33.69:
Question 28:
A sample space consists of 9 elementary events E_{1}, E_{2}, E_{3}, ..., E_{9} whose probabilities are
P(E_{1}) = P(E_{2}) = 0.08, P(E_{3}) = P(E_{4}) = P(E_{5}) = 0.1, P(E_{6}) = P(E_{7}) = 0.2, P(E_{8}) = P(E_{9}) = 0.07
Suppose A = {E_{1}, E_{5}, E_{8}}, B = {E_{2}, E_{5}, E_{8}, E_{9}} [NCERT EXEMPLAR]
(i) Compute P(A), P(B) and P(A ∩ B).
(ii) Using the addition law of probability, find P(A ∪ B).
(iii) List the composition of the event A ∪ B, and calculate P(A ∪ B) by addting the probabilities of elementary events.
(iv) Calculate $\mathrm{P}\left(\overline{)B}\right)$ from P(B), also calculate $\mathrm{P}\left(\overline{)B}\right)$ directly from the elementary events of $\overline{)B}$.
Answer:
Disclaimer: The solution of the problem is provided by taking P(E_{5}) = 0.1. This information is missing in the question as given in the book.
Let S be the sample space of the elementary events.
S = {E_{1}, E_{2}, E_{3}, ..., E_{9}}
Given:
A = {E_{1}, E_{5}, E_{8}}
B = {E_{2}, E_{5}, E_{8}, E_{9}}
P(E_{1}) = P(E_{2}) = 0.08, P(E_{3}) = P(E_{4}) = P(E_{5}) = 0.1, P(E_{6}) = P(E_{7}) = 0.2, P(E_{8}) = P(E_{9}) = 0.07
(i)
P(A) = P(E_{1}) + P(E_{5}) + P(E_{8}) = 0.08 + 0.1 + 0.07 = 0.25
P(B) = P(E_{2}) + P(E_{5}) + P(E_{8}) + P(E_{9}) = 0.08 + 0.1 + 0.07 + 0.07 = 0.32
Now, A ∩ B = {E_{5}, E_{8}}
∴ P(A ∩ B) = P(E_{5}) + P(E_{8}) = 0.1 + 0.07 = 0.17
(ii) By the addition law of probability, we have
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
= 0.25 + 0.32 − 0.17
= 0.40
(iii)
A = {E_{1}, E_{5}, E_{8}}
B = {E_{2}, E_{5}, E_{8}, E_{9}}
Now, A ∪ B = {E_{1}, E_{2}, E_{5}, E_{8}, E_{9}}
∴ P(A ∪ B) = P(E_{1}) + P(E_{2}) + P(E_{5}) + P(E_{8}) + P(E_{9})
= 0.08 + 0.08 + 0.1 + 0.07 + 0.07
= 0.40
(iv)
$\mathrm{P}\left(\overline{)B}\right)=1-\mathrm{P}\left(B\right)=1-0.32=0.68$ [From (i)]
Also, we know that
$\overline{)B}$ = S − B = {E_{1}, E_{3}, E_{4}, E_{6}, E_{7}}
∴ $\mathrm{P}\left(\overline{)B}\right)$ = P(E_{1}) + P(E_{3}) + P(E_{4}) + P(E_{6}) + P(E_{7})
= 0.08 + 0.1 + 0.1 + 0.2 + 0.2
= 0.68
Page No 33.70:
Question 1:
Three numbers are chosen at random from numbers 1 to 30. Write the probability that the chosen numbers are consecutive.
Answer:
Let S be the sample space associated with three numbers that are chosen at random from the numbers 1 to 30.
∴ n(S) = ^{30}C_{3} = $\frac{30!}{27!\times 3!}=\frac{30\times 29\times 28\times 27!}{27!\times 3\times 2\times 1}=5\times 29\times 28=4060$
Let E be the favourable number of elementary events that the chosen numbers are consecutive.
E = {(1, 2, 3), (2, 3, 4), (3, 4, 5),..., (27, 28, 29), (28, 29, 30)}
i.e. n(E) = ^{28}C_{1} = 28
Hence, required probability = $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{28}{4060}=\frac{1}{145}$
Page No 33.7:
Question 11:
A coin is tossed. If it shows tail, we draw a ball from a box which contains 2 red 3 black balls; if it shows head, we throw a die. Find the sample space of this experiment.
Answer:
The box contains two red balls and three black balls.
Let us denote the red balls as R_{1} and R_{2} and the three black balls as B_{1}, B_{2} and B_{3}.
The sample space of this experiment is given by
S = {(T, R_{1}), (T, R_{2}), (T, B_{1}), (T, B_{2}), (T, B_{3}), (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
Page No 33.7:
Question 12:
A coin is tossed repeatedly until a tail comes up for the first time. Write the sample space for this experiment.
Answer:
In this experiment, a tail (T) may come up on the first throw, the second throw, the third throw and so on, until T is obtained.
This process continues indefinitely.
Hence, the sample space of this experiment is given by
S = {T, HT, HHT, HHHT, HHHHT, HHHHHT, ...}
Page No 33.7:
Question 13:
A box contains 1 red and 3 black balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.
Answer:
The box contains one red ball and three black balls.
Let us denote the red ball as R and the three black balls as B_{1}, B_{2} and B_{3}.
The sample space of this experiment is given by
S = {(R, B_{1}), (R, B_{2}), (R, B_{3}), (B_{1}, R), (B_{1}, B_{2}), (B_{1}, B_{3}), (B_{2}, B_{1}), (B_{2}, B_{3}), (B_{2}, R), (B_{3}, R), (B_{3}, B_{1}), (B_{3}, B_{2})}
Page No 33.7:
Question 14:
A pair of dice is rolled. If the outcome is a doublet, a coin is tossed. Determine the total number of elementary events associated to this experiment.
Answer:
If a pair of dices is thrown simultaneously, then all possible outcomes = 6 × 6 = 36
The set of these outcomes is the sample space, which is given by
S = { (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Again, if the outcome is a doublet, then a coin is tossed.
Now, we have the following events:
{(1, 1, H), (2, 2, H), (3, 3, H), (4, 4, H), (5, 5, H), (6, 6, H),
(1, 1, T), (2, 2, T), (3, 3, T), (4, 4, T), (5, 5, T), (6, 6, T)}
Total number of events when the outcome is a doublet = 6 x 2 = 12
Hence, the total number of elementary events associated with this experiment = (36 − 6) + 12 = 42
Page No 33.7:
Question 15:
A coin is tossed twice. If the second draw results in a head, a die is rolled. Write the sample space for this experiment.
Answer:
When a coin is tossed, the possible outcomes are head (H) and tail (T).
If a coin is tossed twice, the possible outcomes are given by
S = {HH, HT, TH, TT}
Again, if the second draw results in a head, then a dice is rolled. The events associated with this experiment is given by
A = {(HH, 1), (HH, 2), (HH, 3), (HH, 4), (HH, 5), (HH, 6), (TH, 1), (TH, 2), (TH, 3), (TH, 4), (TH, 5), (TH, 6)}
Hence, the sample space for this experiment is given by
(S ∪ A) = {(HH, 1), (HH, 2), (HH, 3), (HH, 4), (HH, 5), (HH, 6), (TH, 1),
(TH, 2), (TH, 3), (TH, 4), (TH, 5), (TH, 6), (HT), (TT)}
Page No 33.7:
Question 16:
A bag contains 4 identical red balls and 3 identical black balls. The experiment consists of drawing one ball, then putting it into the bag and again drawing a ball. What are the possible outcomes of the experiment?
Answer:
A bag contains four identical red balls (R) and three identical black balls (B).
The sample space S of drawing one ball with replacement and then again drawing a ball is given by
S = {RR, RB, BR, BB}
Page No 33.7:
Question 17:
In a random sampling three items are selected from a lot. Each item is tested and classified as defective (D) or non-defective (N). Write the sample space of this experiment.
Answer:
Three items are to be selected at random from a lot.
Each item in the lot is tested and classified as defective (D) or non-defective (N).
The sample space of this experiment is given by
S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}
Page No 33.7:
Question 18:
An experiment consists of boy-girl composition of families with 2 children.
(i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?
(ii) What is the sample space if we are interested in the number of boys in a family?
Answer:
(i) When the order of the birth of a girl or a boy is considered, the sample space is given by
S = {(G_{1}, G_{2}), (G_{1}, B_{2}), (B_{1} ,G_{2}), (B_{1}, B_{2})}
(ii) Since the maximum number of children in each family is two, a family can either have two boys or one boy or no boy.
Hence, the required sample space is given by
S = {0, 1, 2}
Page No 33.7:
Question 19:
There are three coloured dice of red, white and black colour. These dice are placed in a bag. One die is drawn at random from the bag and rolled its colour and the number on its uppermost face is noted. Describe the sample space for this experiment.
Answer:
A dice has six faces that are numbered from 1 to 6, with one number on each face.
Let us denote the red, white and black dices as R, W, and B, respectively.
Accordingly, when a dice is selected and then rolled, the sample space is given by
S = {(R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6),
(B, 1), (B, 2), (B, 3), (B, 4), (B, 5), (B, 6),
(W, 1), (W, 2), (W, 3), (W, 4), (W, 5), (W, 6)}
Page No 33.7:
Question 20:
2 boys and 2 girls are in room P and 1 boy 3 girls are in room Q. Write the sample space for the experiment in which a room is selected and then a person.
Answer:
Let us denote two boys and two girls in room P as B_{1}, B_{2} and G_{1}, G_{2}, respectively.
Let us denote one boy and three girls in room Q as B_{3} and G_{3}, G_{4}, G_{5}_{, }respectively.
Accordingly, the required sample space is given by S = {(P, B_{1}), (P, B_{2}), (P, G_{1}), (P, G_{2}) (Q, B_{3}), (Q, G_{3}), (Q, G_{4}),
(Q, G_{5})}.
Page No 33.7:
Question 21:
A bag contains one white and one red ball. A ball is drawn from the bag. If the ball drawn is white it is replaced in the bag and again a ball is drawn. Otherwise, a die is tossed. Write the sample space for this experiment.
Answer:
A bag contains one white ball (W) and one red ball (R).
When one ball is drawn, it will be either W or R.
The sample space of drawing one white ball with replacement and then again drawing a ball is {(W, W), (W, R)}.
Again, if red ball is drawn, a dice is rolled.
The sample space associated with this experiment is given by {(R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6)}.
Hence, the sample space S for this experiment is S = {(W, W), (W, R), (R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6)}.
Page No 33.7:
Question 22:
A box contains 1 white and 3 identical black balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.
Answer:
It is given that the box contains one white ball and three identical black balls.
Let us denote the white ball with W and a black ball with B.
When two balls are drawn at random in succession without replacement, the sample space for this experiment will be given by
S = {WB, BW, BB}
Page No 33.7:
Question 23:
An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.
Answer:
A dice has six faces that are numbered from 1 to 6, with one number on each face.
Here, 2, 4 and 6 are even numbers, while 1, 3 and 5 are odd numbers.
A coin has two faces: a head (H) and a tail (T).
Hence, the sample space of this experiment is given by
S = { (2, H), (2, T), (4, H), (4, T), (6, H), (6, T), (1, HH), (1, HT), (1,TH), (1, TT), (3, HH), (3, HT), (3, TH), (3, TT),
(5, HH), (5, HT), (5, TH), (5, TT) }
Page No 33.7:
Question 24:
A die is thrown repeatedly until a six comes up. What is the sample space for this experiment.
Answer:
In this experiment, 6 may come up on the first throw, the second throw, the third throw and so on, until it is obtained.
Hence, the sample space of this experiment is given by
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), (1, 3, 6), (1, 4, 6), (1, 5, 6), (2, 1, 6), (2, 2, 6), (2, 3, 6) …}
Page No 33.71:
Question 2:
n (≥ 3) persons are sitting in a row. Two of them are selected. Write the probability that they are together.
Answer:
It is given that n (≥ 3) persons are seated in a row and two persons are selected.
∴ Total number of elementary event = n(S) = ^{n}C_{2}
Let E be the event associated with the experiment that two persons are together.
∴ n(E) = ^{n }^{-1}C_{1}
Thus, required probability = P(E) = $\frac{n\left(E\right)}{n\left(S\right)}$
=$\frac{{}^{n-1}C_{1}}{{}^{n}C_{2}}\phantom{\rule{0ex}{0ex}}$
=$\frac{\left(n-1\right)}{{\displaystyle \frac{n\left(n-1\right)}{2}}}=\frac{2\left(n-1\right)}{n\left(n-1\right)}=\frac{2}{n}$
Page No 33.71:
Question 3:
A single letter is selected at random from the word 'PROBABILITY'. What is the probability that it is a vowel?
Answer:
There are 11 letters in the word 'PROBABILITY'.
i.e. n(S) = 11
There are 4 vowels (O, A, I, I) in the given word.
i.e. n(vowel) = 4
∴ Probability (vowel) = $\frac{4}{11}$
Page No 33.71:
Question 4:
What is the probability that a leap year will have 53 Fridays or 53 Saturdays?
Answer:
We know that a leap year has 366 days (i.e. 7 $\times $ 52 + 2) = 52 weeks and 2 extra days
The sample space for these two extra days are as follows:
S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
There are 7 cases.
i.e. n(S) = 7
Let E be the event that a leap year has 53 Fridays or 53 Saturdays.
Then E = { (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
i.e. n(E) = 3
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{7}$
Hence, the probability that a leap year has 53 Fridays or 53 Saturdays is $\frac{3}{7}.$
Page No 33.71:
Question 5:
Three dice are thrown simultaneously. What is the probability of getting 15 as the sum?
Answer:
If three dices are thrown simultaneously, then the number of all the possible outcomes are 6^{3} = 216.
∴ Total number of possible outcome = n(S) = 216
Let A be the event of getting a sum of 15 when three dices are thrown simultaneously.
The favourable outcomes are as follows:
A = {(3,6 , 6), (4, 6, 5), (5, 6, 4), (6, 6, 3), (6, 3, 6), (6, 4, 5), (6, 5, 4), (4, 5, 6), (5, 5, 5), (5, 4, 6)}
i.e. number of favourable outcomes = n(A) = 10
Hence, required probability = P (getting a sum of 15) = $\frac{10}{216}$
Page No 33.71:
Question 6:
If the letters of the word 'MISSISSIPPI' are written down at random in a row, what is the probability that four S's come together.
Answer:
There are 11 letters in the word ‘MISSISSIPPI’ which can be arranged in 11! ways.
Number of the letter S = 4
Let us consider the four S's in the given word as one letter.
So, when the four letters are clubbed together, we have (SSSS) MIIIPPI. We can arrange eight letters in a row in 8! ways.
Also, the four S's can be arranges in 4! ways.
Hence, required probability = $\frac{8!\times 4!}{11!}=\frac{8!\times 4\times 3\times 2}{11\times 10\times 9\times 8!}=\frac{4\times 3\times 2}{11\times 10\times 9}=\frac{4}{165}$
Page No 33.71:
Question 7:
What is the probability that the 13th days of a randomly chosen month is Friday?
Answer:
Probability of any chosen month out of 12 months = $\frac{1}{12}$
There are seven possible ways in which a month can start and it will be a Friday on the 13^{th} day if the first day of the month is Sunday.
So, its probability = $\frac{1}{7}$
Thus, required probability = $\frac{1}{12}\times \frac{1}{7}=\frac{1}{84}$
Hence, the probability that the 13^{th} day of a randomly chosen month is Friday = $\frac{1}{84}$
Page No 33.71:
Question 8:
Three of the six vertices of a regular hexagon are chosen at random. What is the probability that the triangle with these vertices is equilateral.
Answer:
We choose three vertices out of 6 in ^{6}C_{3} = 20 ways.
∴ Total number of elementary events = n(S) = 20
Number of ways of choosing an equilateral triangle = 2
i.e. A_{1}A_{3}A_{5} or A_{2}A_{4}A_{6}
∴ Total number of favourable events = n(E) = 2
Hence, required probability = $\frac{2}{20}=\frac{1}{10}$
Page No 33.71:
Question 9:
If E and E_{2} are independent evens, write the value of P $\left(({E}_{1}\cup {E}_{2})\cap (\overline{)E}\cap {\overline{)E}}_{2})\right)$.
Answer:
$P\left(\left({E}_{1}\cup {E}_{2}\right)\cap \left(\overline{{E}_{1}}\cap \overline{{E}_{2}}\right)\right)=P\left({E}_{1}\cap \left(\overline{{E}_{1}}\cap \overline{{E}_{2}}\right)\right)\cup \left({E}_{2}\cap \left(\overline{{E}_{1}}\cap \overline{{E}_{2}}\right)\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$=P\left(\varphi \cup \varphi \right)=0$
Page No 33.71:
Question 10:
If A and B are two independent events such that $P(A\cap B)=\frac{1}{6}\mathrm{and}P(\overline{)A}\cap \overline{)B})=\frac{1}{3},$ then write the values of P (A) and P (B).
Answer:
$P\left(\overline{A}\cap \overline{)B}\right)=1-P\left(A\cup B\right)\phantom{\rule{0ex}{0ex}}$
$\Rightarrow P\left(A\cup B\right)=1-P\left(\overline{A}\cap \overline{)B}\right)=1-\frac{1}{3}=\frac{2}{3}$
By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) $-$ P (A ∩ B)
∴ P(A) + P (B) = P (A ∪ B) + P (A ∩ B)
= $\frac{2}{3}+\frac{1}{6}=\frac{4+1}{6}=\frac{5}{6}$
Thus, P(A) + P (B) = $\frac{5}{6}$ ...(i)
Again, P (A ∩ B) = P(A) × P(A) = $\frac{1}{6}$
By formula, we have:
{P(A) − P (B)}^{2} = {P(A) + P (B)}^{2} − 4 × P(A) × P(B)
= ${\left(\frac{5}{6}\right)}^{2}-\frac{4}{6}=\frac{25}{36}-\frac{4}{6}=\frac{25-24}{36}=\frac{1}{36}$
∴ P(A) − P(B) = $\frac{1}{6}$ ...(ii)
From (i) and (ii), we get:
2P(A) = 1
Hence, P(A) = $\frac{1}{2}$ and P(B) = $\frac{1}{3}$
Page No 33.71:
Question 1:
One card is drawn from a pack of 52 cards. The probability that it is the card of a king or spade is
(a) 1/26
(b) 3/26
(c) 4/13
(d) 3/13
Answer:
(c) 4/13
If A and B denote the events of drawing a king and a spade card, respectively, then event A consists of four sample points, whereas event B consists of 13 sample points.
Thus, $P\left(A\right)=\frac{4}{52}$ and $P\left(B\right)=\frac{13}{52}$
The compound event (A ∩ B) consists of only one sample point, king of spade.
So, $P\left(A\cap B\right)=\frac{1}{52}$
By addition theorem , we have:
P (A ∪ B) = P(A) + P (B) − P (A ∩ B)
= $\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$
Hence, the probability that the card drawn is either a king or a spade is given by $\frac{4}{13}$.
Page No 33.71:
Question 2:
Two dice are thrown together. The probability that at least one will show its digit greater than 3 is
(a) 1/4
(b) 3/4
(c) 1/2
(d) 1/8
Answer:
(b) 3/4
When two dice are thrown, there are (6 × 6) = 36 outcomes.
The set of all these outcomes is the sample space, given by
S = (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
i.e. n(S) = 36
Let E be the event of getting at least one digit greater than 3.
Then E = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
∴ n(E) = 27
Hence, required probability = $\frac{27}{36}=\frac{3}{4}$
Page No 33.71:
Question 3:
Two dice are thrown simultaneously. The probability of obtaining a total score of 5 is
(a) 1/18
(b) 1/12
(c) 1/9
(d) none of these
Answer:
(c) 1/9
When two dice are thrown, there are (6 × 6) = 36 outcomes.
The set of all these outcomes is the sample space given by
S = (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
i.e. n(S) = 36
Let E be the event of getting a total score of 5.
Then E = {(1, 4), (2, 3), (3, 2), (4, 1)}
∴ n(E) = 4
Hence, required probability = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{4}{36}=\frac{1}{9}$
Page No 33.71:
Question 4:
Two dice are thrown simultaneously. The probability of obtaining total score of seven is
(a) 5/36
(b) 6/36
(c) 7/36
(d) 8/36
Answer:
(b) 6/36
When two dices are thrown, there are (6 × 6) = 36 outcomes.
The set of all these outcomes is the sample space given by
S = (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
∴ n(S) = 36
Let E be the event of getting a total score of 7.
Then E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
∴ n(E) = 6
Hence, required probability = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{6}{36}$
Page No 33.71:
Question 5:
The probability of getting a total of 10 in a single throw of two dices is
(a) 1/9
(b) 1/12
(c) 1/6
(d) 5/36
Answer:
(b) 1/12
When two dices are thrown, there are (6 × 6) = 36 outcomes.
The set of all these outcomes is the sample space, given by
S = (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
i.e. n(S) = 36
Let E be the event of getting a total score of 10.
Then E = {(4, 6), (5, 5), (6, 4)}
∴ n(E) = 3
Hence, required probability = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{36}=\frac{1}{12}$
Page No 33.71:
Question 6:
A card is drawn at random from a pack of 100 cards numbered 1 to 100. The probability of drawing a number which is a square is
(a) 1/5
(b) 2/5
(c) 1/10
(d) none of these
Answer:
(c) 1/10
Clearly, the sample space is given by
S = {1, 2, 3, 4, 5... 97, 98, 99, 100}
∴ n(S) = 100
Let E = event of getting a square.
Then E = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
∴ n(E) = 10
Hence, required probability = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{100}=\frac{1}{10}$
Page No 33.71:
Question 7:
A bag contains 3 red, 4 white and 5 blue balls. All balls are different. Two balls are drawn at random. The probability that they are of different colour is
(a) 47/66
(b) 10/33
(c) 1/3
(d) 1
Answer:
(a) 47/66
Out of 12 balls, two balls can be drawn in ^{12}C_{2} ways.
∴ Total number of elementary events, n(S) = ^{12}C_{2} = 66
We observe that at least one ball of each colour can be drawn in one of the following mutually exclusive ways:
(i) 1 red and 1 white
(ii) 1 red and 1 blue
(iii) 1 white and 1 blue
Thus, if we define three events A, B and C as follows:
A = drawing 1 red and 1 white
B = drawing 1 red and 1 blue
C = drawing 1 white and 1 blue
then, A, B and C are mutually exclusive events.
∴ Required probability = P(A ∪ B ∪ C)
= P(A) + P(B) + P(C)
= $\frac{{}^{3}C_{1}\times {}^{4}C_{1}}{{}^{12}C_{2}}+\frac{{}^{3}C_{1}\times {}^{5}C_{1}}{{}^{12}C_{2}}+\frac{{}^{4}C_{1}\times {}^{5}C_{1}}{{}^{12}C_{2}}\phantom{\rule{0ex}{0ex}}$
= $\frac{3\times 4}{66}+\frac{3\times 5}{66}+\frac{4\times 5}{66}\phantom{\rule{0ex}{0ex}}$
= $\frac{12}{66}+\frac{15}{66}+\frac{20}{56}=\frac{47}{66}$
Page No 33.71:
Question 8:
Two dice are thrown together. The probability that neither they show equal digits nor the sum of their digits is 9 will be
(a) 13/15
(b) 13/18
(c) 1/9
(d) 8/9
Answer:
(b) 13/18
When two dices are thrown, there are (6 × 6) = 36 outcomes.
The set of all these outcomes is the sample space is given by
S = (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
∴ n(S) = 36
Let E be the event of getting the digits which are neither equal nor give a total of 9.
Then E' = event of getting either a doublet or a total of 9
Thus, E' = {{1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 6), (4, 5), (5, 4), (6, 3)}
i.e. n(E') = 10
P(E') = $\frac{n\left(E\text{'}\right)}{n\left(S\right)}=\frac{10}{36}=\frac{5}{18}$
Hence, required probability P(E) = 1$-$ P(E')
= $1-\frac{5}{18}=\frac{13}{18}$
Page No 33.71:
Question 9:
Four persons are selected at random out of 3 men, 2 women and 4 children. The probability that there are exactly 2 children in the selection is
(a) 11/21
(b) 9/21
(c) 10/21
(d) none of these
Answer:
(c) 10/21
There are nine persons (three men, two women and four children) out of which four persons can be selected in ^{9}C_{4} = 126 ways.
∴ Total number of elementary events = 126
Exactly two children means selecting two children and two other people from three men and two women.
This can be done in ^{4}C_{2} × ^{5}C_{2} ways.
∴ Favourable number of elementary events = ^{4}C_{2} × ^{5}C_{2} = 60
So, required probability = $\frac{60}{126}=\frac{10}{21}$
Page No 33.71:
Question 10:
The probabilities of happening of two events A and B are 0.25 and 0.50 respectively. If the probability of happening of A and B together is 0.14, then probability that neither A nor B happens is
(a) 0.39
(b) 0.25
(c) 0.11
(d) none of these
Answer:
(a) 0.39
Given:
P(A) = 0.25, P(B) = 0.50 and P(A$\cap $B) = 0.14
∴ Required probability = 1 $-$P(A∪B)
= 1 $-$ [P(A) + P(B) $-$ P(A$\cap $B)]
= 1 $-$ [0.25 + 0.50 $-$ 0.14]
= 1 $-$ 0.61 = 0.39
Page No 33.72:
Question 11:
A die is rolled, then the probability that a number 1 or 6 may appear is
(a) 2/3
(b) 5/6
(c) 1/3
(d) 1/2
Answer:
(c) 1/3
Total number of sample space, S = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Let A be the event of getting the number 1 or 6.
A = {1, 6}
i.e. n(A) = 2
Hence, required probability = $\frac{n\left(A\right)}{n\left(S\right)}=\frac{2}{6}=\frac{1}{3}$
Page No 33.72:
Question 12:
Six boys and six girls sit in a row randomly. The probability that all girls sit together is
(a) 1/122
(b) 1/112
(c) 1/102
(d) 1/132
Answer:
(d) $\frac{1}{132}$
Total number of ways in which six boys and six girls can be seated in a row = (12)!
Taking all the six girls as one person, seven persons can be seated in a row in 7! ways. The six girls can be arranged among themselves in 6! ways.
Then number of ways in which six boys and six girls can be seated in a row so that all the girls sit together = 7! × 6!
∴ Required probability = $\frac{7!\times 6!}{\left(12\right)!}=\frac{720}{12\times 11\times 10\times 9\times 8}=\frac{1}{132}$
Page No 33.72:
Question 13:
The probabilities of three mutually exclusive events A, B and C are given by 2/3, 1/4 and 1/6 respectively. The statement
(a) is true
(b) is false
(c) nothing can be said
(d) could be either
Answer:
(b) is false
Since the events A, B and C are mutually exclusive, we have:
$P\left(A\cup B\cup C\right)=\frac{2}{3}+\frac{1}{4}+\frac{1}{6}=\frac{13}{12}>1\phantom{\rule{0ex}{0ex}}$ , which is not possible.
Hence, the given statement is false.
Page No 33.72:
Question 14:
If $\frac{(1-3p)}{2},\frac{(1+4p)}{3},\frac{(1+p)}{6}$ are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of p is
(a) (0, 1)
(b) (−1/4, 1/3)
(c) (0, 1/3)
(d) (0, ∞)
Answer:
(b) (−1/4, 1/3)
P(A) = (1 − 3p)/2
P(B) = (1 + 4p)/3
P(C) = (1 + p)/6
The events are mutually exclusive and exhaustive.
∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 1
⇒ $0\le \mathrm{P}\left(\mathrm{A}\right)\le 1$, $0\le \mathrm{P}\left(\mathrm{B}\right)\le 1$, $0\le \mathrm{P}\left(\mathrm{C}\right)\le 1$
⇒ $0\le \frac{1-3p}{2}\le 1\phantom{\rule{0ex}{0ex}}$, $0\le \frac{1+4p}{3}\le 1\phantom{\rule{0ex}{0ex}}$, $0\le \frac{1+p}{6}\le 1\phantom{\rule{0ex}{0ex}}$
$\Rightarrow -1/3\le p\le \frac{1}{3}\phantom{\rule{0ex}{0ex}}$, ...(i)
$\frac{-1}{4}\le p\le \frac{1}{2}$ ...(ii)
and $-1\le p\le 5$ ...(iii)
The common solution of (i), (ii) and (iii) is $-1/4\le p\le 1/3$.
∴ The set values of p are ($-$1/4 , 1/3)
Page No 33.72:
Question 15:
A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. Two cards are drawn at random. The probability that at least one of them is an ace is
(a) 1/5
(b) 3/16
(c) 9/20
(d) 1/9
Answer:
(c) $\frac{9}{20}$
We have:
P(both are aces) = $\frac{{}^{4}{C}_{2}}{{}^{16}{C}_{2}}=$$\frac{4}{16}\times \frac{3}{15}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}$
P(one is ace) = $\frac{{}^{4}{C}_{1}{\times}^{12}{C}_{1}}{C{{}^{16}}_{2}}=\frac{2}{5}$
∴ P(at least one is ace) = $\frac{1}{20}+\frac{2}{5}=\frac{9}{20}$
Page No 33.72:
Question 16:
If three dice are throw simultaneously, then the probability of getting a score of 5 is
(a) 5/216
(b) 1/6
(c) 1/36
(d) none of these
Answer:
(c) 1/36
When three dice are thrown together, the sample space S associated with the random experiment is given by
S = {(1, 1, 1), (1, 1, 2), (1, 1, 3) ...(6, 6, 5), (6, 6, 6)}
Clearly, total number of elementary events n(S) = 216
Let A be the event of getting a total score of 5.
Then A = { (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1)}
∴ Favourable number of elementary events = 6
i.e. n(A) = 6
Hence, required probability = $\frac{6}{216}=\frac{1}{36}$
Page No 33.72:
Question 17:
One of the two events must occur. If the chance of one is 2/3 of the other, then odds in favour of the other are
(a) 1 : 3
(b) 3 : 1
(c) 2 : 3
(d) 3 : 2
Answer:
(d) 3 : 2
Let P(B) = x
Then, P(A) = $\frac{2x}{3}$
P(A) + P(B) = $x+\frac{2x}{3}=\frac{5x}{3}$
$\Rightarrow \frac{5x}{3}=1\phantom{\rule{0ex}{0ex}}$ (∵ They are exhaustive events)
$\Rightarrow x=\frac{3}{5}\phantom{\rule{0ex}{0ex}}$
Now, $P\left(A\right)=\frac{2}{5}\mathrm{and}P\left(B\right)=\frac{3}{5}$
∴ Odd in favour of B = $\frac{3/5}{1-3/5}=\frac{3}{2}=3:2$
Page No 33.72:
Question 18:
The probability that a leap year will have 53 Fridays or 53 Saturdays is
(a) 2/7
(b) 3/7
(c) 4/7
(d) 1/7
Answer:
(b) 3/7
We know that a leap year has 366 days (i.e. 7 $\times $ 52 + 2) = 52 weeks and 2 extra days
The sample space for these 2 extra days is given below:
S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
There are 7 cases.
∴ n(S) = 7
Let E be the event that the leap year has 53 Fridays or 53 Saturdays.
E = { (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
i.e. n(E) = 3
Hence, the probability that a leap year has 53 Fridays or 53 Saturdays is
Page No 33.72:
Question 19:
A person write 4 letters and addresses 4 envelopes. If the letters are placed in the envelopes at random, then the probability that all letters are not placed in the right envelopes, is
(a) 1/4
(b) 11/24
(c) 15/24
(d) 23/24
Answer:
(d) $\frac{23}{24}$
Total number of ways of placing four letters in 4 envelops = 4! = 24
All the letters can be dispatched in the right envelops in only one way. Therefore, the probability that all the letters are placed in the right envelops is $\frac{1}{24}$.
Hence, probability that all the letters are not placed in the right envelops = $1-\frac{1}{24}=\frac{23}{24}$
Page No 33.72:
Question 20:
A and B are two events such that P (A) = 0.25 and P (B) = 0.50. The probability of both happening together is 0.14. The probability of both A and B not happening is
(a) 0.39
(b) 0.25
(c) 0.11
(d) none of these
Answer:
(a) 0.39
P (A) = 0.25 and P (B) = 0.50
P(A$\cap $B) = 0.14
∴ Required probability = 1 $-$ P(A∪B)
= 1 $-$ [P(A) + P(B) $-$ P(A$\cap $B)]
= 1 $-$ [0.25 + 0.50 $-$ 0.14]
= 1 $-$ 0.61 = 0.39
Page No 33.72:
Question 21:
If the probability of A to fail in an examination is $\frac{1}{5}$ and that of B is $\frac{3}{10}$. Then, the probability that either A or B fails is
(a) 1/2
(b) 11/25
(c) 19/50
(d) none of these
Answer:
(c) 19/50
Given:
P(A) = $\frac{1}{5}$
∴ P(A') = $1-\frac{1}{5}=\frac{4}{5}$
P(B) = $\frac{3}{10}$
∴ P(B') = $1-\frac{3}{10}=\frac{7}{30}$
Hence, required probability = P(A∩ B') + P(A'∩ B)
$=\frac{1}{5}\times \frac{7}{10}+\frac{4}{5}\times \frac{3}{10}\phantom{\rule{0ex}{0ex}}=\frac{7}{50}+\frac{12}{50}\phantom{\rule{0ex}{0ex}}=\frac{19}{50}$
Page No 33.72:
Question 22:
A box contains 10 good articles and 6 defective articles. One item is drawn at random. The probability that it is either good or has a defect, is
(a) 64/64
(b) 49/64
(c) 40/64
(d) 24/64
Answer:
(a) $\frac{64}{64}$
The answer is one, because the article would be either good or defective as per the question.
Hence, the only option is $\frac{64}{64}=1$.
Page No 33.72:
Question 23:
Three integers are chosen at random from the first 20 integers. The probability that their product is even is
(a) 2/19
(b) 3/29
(c) 17/19
(d) 4/19
Answer:
(c) 17/19
Number of ways in which we can choose three distinct integers from 20 integers = ^{20}C_{3} = 1140
We know that, if we take three odd numbers, there product will always be an odd number.
Out of 20 consecutive integers, 10 are even and 10 are odd integers.
Number of ways in which we can choose three distinct odd integers from 10 odd integers= ^{10}C_{3} = 120
P(product is even) = 1 $-$ P(product is odd)
= $1-\frac{120}{1140}=\frac{1140-120}{1140}=\frac{1020}{1140}=\frac{17}{19}$
Page No 33.72:
Question 24:
Out of 30 consecutive integers, 2 are chosen at random. The probability that their sum is odd, is
(a) 14/29
(b) 16/29
(c) 15/29
(d) 10/29
Answer:
(c) 15/29
The total number of ways in which two integers can be chosen from the given 30 integers is ^{30}C_{2}.
The sum of the selected numbers is odd if exactly one of them is even or odd.
∴ Favourable number of outcomes = ^{15}C_{1} × ^{15}C_{1}
Hence, required probability = $\frac{{}^{15}C_{1}\times {}^{15}C_{1}}{{}^{30}C_{2}}=\frac{15}{29}$
Page No 33.72:
Question 25:
A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected randomwise, the probability that it is black or red ball is
(a) 1/3
(b) 1/4
(c) 5/12
(d) 2/3
Answer:
(d) 2/3
Out of 12 balls, one ball can be drawn in ^{12}C_{1} ways.
∴ Total number of elementary events = ^{12}C_{1} = 12
Out of fivne black balls, one black ball can be chosen in ^{5}C_{1} = 5 ways.
Out of three red balls, one red ball can be chosen in ^{3}C_{1} = 3 ways.
∴ Favourable number of events = 5 + 3 = 8
Hence, required probability = $\frac{8}{12}=\frac{2}{3}$
Page No 33.72:
Question 26:
Two dice are thrown simultaneously. The probability of getting a pair of aces is
(a) 1/36
(b) 1/3
(c) 1/6
(d) none of these
Answer:
(a) 1/36
When two dice are thrown simultaneously, the sample space associated with the random experiment is given by:
S = {(1, 1), (1, 2), (1, 3) ...(6, 4), (6, 5), (6, 6)}
Clearly, total number of elementary events = 36
Let A be the event of getting a pair of aces.
Then A = {(1, 1)}
∴ n(A) = 1
Hence, required probability = $\frac{n\left(A\right)}{n\left(S\right)}=\frac{1}{36}$
Page No 33.72:
Question 27:
An urn contains 9 balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of the same colour is
(a) 5/84
(b) 3/9
(c) 3/7
(d) 7/17
Answer:
(a) 5/84
Three balls can be drawn randomly from nine balls in ^{9}C_{3} = 84 ways.
Three balls cannot be red as there are only two red balls.
Three balls of the same colour can be drawn in the following ways :
3 blue out of a total of 3 blue balls.
The probability for which is $\frac{{}^{3}C_{3}}{84}=\frac{1}{84}$.
3 black out of a total of 4 black balls.
The probability for which is $\frac{{}^{4}C_{3}}{84}=\frac{4}{84}$.
Hence, required probability = $\frac{1}{84}+\frac{4}{84}=\frac{5}{84}$
Page No 33.72:
Question 28:
Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first, then the probability of all 5 persons leaving at different floor is
(a) $\frac{{}^{7}P_{5}}{{7}^{5}}$
(b) $\frac{{7}^{5}}{{}^{7}P_{5}}$
(c) $\frac{6}{{}^{6}P_{5}}$
(d) $\frac{{}^{5}P_{5}}{{5}^{5}}$
Answer:
(a) $\frac{{}^{7}P_{5}}{{7}^{5}}$
Since, it is an eight-storey building.
So, there are 7 possible options for them in 7 floors in total if ground floor is not considered.
Hence, total possible outcomes = 7× 7× 7 × 7 × 7= 7^{5}
Thus, number of ways in which 5 persons can leave from seven floors differently = ^{7}P_{5}
∴ Required probability = $\frac{{}^{7}P_{5}}{{7}^{5}}$
Page No 33.73:
Question 29:
A box contains 10 good articles and 6 with defects. One item is drawn at random. The probability that it is either good or has a defect is
(a) 64/64
(b) 49/64
(c) 40/64
(d) 24/64
Answer:
(a) $\frac{64}{64}$
Let A be the event of drawing one good article whereas B be the event of drawing one defected article.
Here,
$P\left(A\right)=\frac{10}{10+6}=\frac{10}{16}\mathrm{and}P\left(B\right)=\frac{6}{10+6}=\frac{6}{16}$
The events A and B are mutually exclusive. Thus, the required probability is $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)$.
$\Rightarrow P\left(A\cup B\right)=\frac{10}{16}+\frac{6}{16}=\frac{16}{16}=1$
Hence, the correct option is (a).
Page No 33.73:
Question 30:
A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one item is chosen at random, the probability that it is rusted or is a nail is
(a) 3/16
(b) 5/16
(c) 11/16
(d) 14/16
Answer:
(c) 11/16
If the numbers of nails and nuts are 6 and 10, respectively, then the numbers of rusted nails and rusted nuts are 3 and 5, respectively.
Total number of items = 6 + 10 = 16
Total number of rusted items = 3 + 5 = 8
Total number of ways of drawing one item = ^{16}C_{1}
Let R and N be the events where both the items drawn are rusted items and nails, respectively.
R and N are not mutually exclusive events, because there are 3 rusted nails.
P(either a rusted item or a nail ) = P (R ∪ N)
= P(R) + P (N) $-$ P (R ∩ N)
= $\frac{{}^{6}C_{1}}{{}^{16}C_{1}}+\frac{{}^{8}C_{1}}{{}^{16}C_{1}}-\frac{{}^{3}C_{1}}{{}^{16}C_{1}}\phantom{\rule{0ex}{0ex}}$
= $\frac{6}{16}+\frac{8}{16}-\frac{3}{16}=\frac{11}{16}$
Page No 33.73:
Question 31:
If S is the sample space and P(A) = $\frac{1}{3}$ P(B) and S = A ∪ B, where A and B are two mutually exclusive events, then P (A) =
(a) 1/4
(b) 1/2
(c) 3/4
(d) 3/8
Answer:
(a) 1/4
Let P(B) = p
Then P(A) = $\frac{1}{3}p$
Since A and B are two mutually exclusive events, we have:
A ∪ B = S
⇒ P (A∪B) = P (S)
⇒ P (A∪B) = 1
⇒ P (A) + P (B) = 1
⇒ $\frac{1}{3}p+p=1\phantom{\rule{0ex}{0ex}}$
$\Rightarrow \frac{4p}{3}=1\phantom{\rule{0ex}{0ex}}\Rightarrow p=\frac{3}{4}$
∴ P (A) = $\frac{1}{3}p=\frac{1}{3}\times \frac{3}{4}=\frac{1}{4}$
Page No 33.73:
Question 32:
One mapping is selected at random from all mappings of the set A = {1, 2, 3, ..., n} into itself. The probability that the mapping selected is one to one is
(a) $\frac{1}{{n}^{n}}$ (b) $\frac{1}{n!}$ (c) $\frac{\left(n-1\right)!}{{n}^{n-1}}$ (d) None of these
Answer:
Number of ways to map 1st element in set A = n
Number of ways to map 2nd element in set A = n and so on
∴ Total number of mapping from set A to itself = $n\times n\times ...\times n$ (n times) = ${n}^{n}$
For one to one mapping,
Number of ways to map 1st element in set A = n
Number of ways to map 2nd element in set A = n −1
Number of ways to map 3rd element in set A = n − 2
. . . . . . . .
. . . . . . . .
Number of ways to map nth element in set A = 1
Total number of one to one mappings from set A to itself = $n\times \left(n-1\right)\times \left(n-2\right)\times ...\times 1=n!$
∴ Required probability = $=\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{one}\mathrm{to}\mathrm{one}\mathrm{mappings}\mathrm{from}\mathrm{set}A\mathrm{to}\mathrm{itself}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{mappings}\mathrm{from}\mathrm{set}A\mathrm{to}\mathrm{itself}}=\frac{n!}{{n}^{n}}=\frac{\left(n-1\right)!}{{n}^{n-1}}$
Hence, the correct answer is option (c).
Page No 33.73:
Question 33:
If A, B, C are three mutually exclusive and exhaustive events of an experiment such that 3 P(A) = 2 P(B) = P(C), then P(A) is equal to
(a) $\frac{1}{11}$ (b) $\frac{2}{11}$ (c) $\frac{5}{11}$ (d) $\frac{6}{11}$
Answer:
Let 3 P(A) = 2 P(B) = P(C) = p. Then,
P(A) = $\frac{p}{3}$, P(B) = $\frac{p}{2}$ and P(C) = p
It is given that A, B, C are three mutually exclusive and exhaustive events.
∴ P(A) + P(B) + P(C) = 1 [P(A ∩ B) = P(B ∩ C) = P(C ∩ A) = P(A ∩ B ∩ C) = 0 and P(A ∪ B ∪ C) = 1]
$\Rightarrow \frac{p}{3}+\frac{p}{2}+p=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{11p}{6}=1\phantom{\rule{0ex}{0ex}}\Rightarrow p=\frac{6}{11}$
$\therefore \mathrm{P}\left(A\right)=\frac{p}{3}=\frac{{\displaystyle \frac{6}{11}}}{3}=\frac{2}{11}$
Hence, the correct answer is option (b).
Page No 33.73:
Question 34:
If A and B are mutually exclusive events then
(a) $\mathrm{P}\left(A\right)\le \mathrm{P}\left(\overline{)B}\right)$ (b) $\mathrm{P}\left(A\right)\ge \mathrm{P}\left(\overline{)B}\right)$ (c) $\mathrm{P}\left(A\right)<\mathrm{P}\left(\overline{)B}\right)$ (d) None of these
Answer:
It is given that A and B are mutually exclusive events.
∴ P(A ∩ B) = 0 .....(1)
We know that,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
⇒ P(A ∪ B) = P(A) + P(B) [From (1)]
⇒ P(A) + P(B) ≤ 1 [P(A ∪ B) ≤ 1]
⇒ P(A) ≤ 1 − P(B) = P($\overline{)B}$)
∴ P(A) ≤ P($\overline{)B}$)
Hence, the correct answer is option (a).
Page No 33.73:
Question 35:
If P(A ∪ B) = P(A ∩ B) for any two events A and B, then
(a) P(A) = P(B) (b) P(A) > P(B) (c) P(A) < P(B) (d) None of these
Answer:
We know that
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
⇒ P(A ∩ B) = P(A) + P(B) − P(A ∩ B) [P(A ∪ B) = P(A ∩ B)]
⇒ P(A) − P(A ∩ B) + P(B) − P(A ∩ B) = 0 .....(1)
But,
P(A) − P(A ∩ B) ≥ 0
P(B) − P(A ∩ B) ≥ 0
⇒ P(A) − P(A ∩ B) + P(B) − P(A ∩ B) ≥ 0 .....(2)
From (1) and (2), we have
P(A) − P(A ∩ B) = 0 and P(B) − P(A ∩ B) = 0
⇒ P(A) = P(A ∩ B) and P(B) = P(A ∩ B)
⇒ P(A) = P(B)
Hence, the correct answer is option (a).
Page No 33.73:
Question 36:
Three numbers are chosen from 1 to 20. The probability that they are not consecutive is
(a) $\frac{186}{190}$ (b) $\frac{187}{190}$ (c) $\frac{188}{190}$ (d) $\frac{18}{{}^{20}\mathrm{C}_{3}}$
Answer:
Number of ways to choose three numbers from 1 to 20 = ${}^{20}\mathrm{C}_{3}$ = 1140
Now, the set of three consecutive numbers from 1 to 20 are (1, 2, 3), (2, 3, 4), (3, 4, 5), ...., (18, 19, 20).
So, the number of ways to choose three numbers from 1 to 20 such that they are consecutive is 18.
P(three numbers choosen are consecutive) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{ways}\mathrm{to}\mathrm{choose}\mathrm{three}\mathrm{consecutive}\mathrm{numbers}\mathrm{from}1\mathrm{to}20}{\mathrm{Number}\mathrm{of}\mathrm{ways}\mathrm{to}\mathrm{choose}\mathrm{three}\mathrm{numbers}\mathrm{from}1\mathrm{to}20}=\frac{18}{{}^{20}\mathrm{C}_{3}}=\frac{18}{1140}=\frac{3}{190}$
∴ P(three numbers choosen are not consecutive) = 1 − P(three numbers choosen are consecutive) = $1-\frac{3}{190}=\frac{187}{190}$
Hence, the correct answer is option (b).
Page No 33.73:
Question 37:
6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is
(a) $\frac{1}{432}$ (b) $\frac{12}{431}$ (c) $\frac{1}{132}$ (d) None of these
Answer:
Total number of ways in which 6 boys and 6 girls can sit in a row = 12!
Consider 6 girls as one group, then 6 boys and one group can arrange in 7! ways.
Now, 6 girls in the group can arrange among themselves in 6!.
So, the number of ways in which all the girls sit together is 7! × 6!.
∴ P(all girls sit together) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{ways}\mathrm{in}\mathrm{which}\mathrm{all}\mathrm{girls}\mathrm{sit}\mathrm{together}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{in}\mathrm{which}6\mathrm{boys}\mathrm{and}6\mathrm{girls}\mathrm{sit}\mathrm{in}\mathrm{a}\mathrm{row}}=\frac{7!6!}{12!}=\frac{6\times 5\times 4\times 3\times 2\times 1}{12\times 11\times 10\times 9\times 8}=\frac{1}{132}$
Hence, the correct answer is option (c).
Page No 33.73:
Question 38:
Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is
(a) $\frac{1}{5}$ (b) $\frac{4}{5}$ (c) $\frac{1}{30}$ (d) $\frac{5}{9}$
Answer:
The given digits are 0, 2, 3 and 5.
_____ | _____ | _____ | _____ |
Thousands | Hundreds | Tens | Ones |
Now, there are 3 ways to fill the thousands place (0 cannot occupy the thousands place), 3 ways to fill the hundreds place, 2 ways to fill the tens place and 1 way to fill the ones place.
Total number of four digit numbers formed = 3 × 3 × 2 × 1 = 18
We know that a number is divisible by 5 if it ends in 0 or 5.
When 0 is at the ones place,
Number of four digits numbers divisible by 5 formed = 3 × 2 × 1 = 6
When 5 is at the ones place,
Number of four digits numbers divisible by 5 formed = 2 × 2 × 1 = 4 (0 cannot occupy the thousands place)
Total number of four digit numbers divisible by 5 = 6 + 4 = 10
∴ P(four digit number formed is divisible by 5) = $\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{four}\mathrm{digit}\mathrm{numbers}\mathrm{divisible}\mathrm{by}5}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{four}\mathrm{digit}\mathrm{numbers}\mathrm{formed}}=\frac{10}{18}=\frac{5}{9}$
Hence, the correct answer is option (d).
Page No 33.73:
Question 39:
If the probability for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is
(a) > 0.5 (b) 0.5 (c) ≤ 0.5 (d) 0
Answer:
Let X and Y be two events given by
X : A fails in an examination
Y : B fails in an examination
P(A fails) = P(X) = 0.2
P(B fails) = P(Y) = 0.3
Now, P(either A or B fails) = P(X ∪ Y)
We know that,
P(X ∪ Y) ≤ P(X) + P(Y) = 0.2 + 0.3 = 0.5
⇒ P(X ∪ Y) ≤ 0.5
∴ P(either A or B fails) ≤ 0.5
Hence, the correct answer is option (c).
Page No 33.73:
Question 40:
Three digit numbers are formed using the digits 0, 2, 4, 6, 8. A number is chosen at random out of these numbers. What is the probability that this number has the same digits?
(a) $\frac{1}{16}$ (b) $\frac{16}{25}$ (c) $\frac{1}{645}$ (d) $\frac{1}{25}$
Answer:
The given digits are 0, 2, 4, 6, 8.
____ | ____ | ____ |
Hundreds | Tens | Ones |
Now, there are 4 ways to fill the hundreds place (0 cannot occupy the hundreds place), 5 ways to fill the tens place and 5 ways to fill the ones place.
Total number of 3 digit numbers formed using the given digits = 4 × 5 × 5 = 100
The three digit numbers formed using given digits that have the same digits are 222, 444, 666 and 888.
Number of 3 digit numbers that have the same digits = 4
∴ P(three digit number formed has the same digits) = $\frac{\mathrm{Number}\mathrm{of}3\mathrm{digit}\mathrm{numbers}\mathrm{that}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{digits}}{\mathrm{Total}\mathrm{number}\mathrm{of}3\mathrm{digit}\mathrm{numbers}\mathrm{formed}\mathrm{using}\mathrm{the}\mathrm{given}\mathrm{digits}}=\frac{4}{100}=\frac{1}{25}$
Hence, the correct answer is option (d).
View NCERT Solutions for all chapters of Class 11