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Page No 927:

Question 1:

Fill in the blanks:
(i) The probability of an impossible event is ....... .
(ii) The probability of a sure event is ........ .
(iii) For any event E, P(E) + P (not E) = ........ .
(iv) The probability of a possible but not a sure event lies between ......... and ........ .
(v) The sum of probabilities of all the outcomes of an experiment is ....... .

Answer:

(i) 0
(ii) 1
(iii) 1
(iv) 0, 1
(v) 1

Page No 927:

Question 2:

A coin is tossed once. What is the probability of getting a tail?

Answer:

When a coin is tossed once, the possible outcomes are H and T.
Total number of possible outcomes = 2
Favourable outcome = 1
∴ Probability of getting a tail =  (T) = Number of favourable outcomesTotal number of possible outcomes = 12

Page No 927:

Question 3:

Two coins are tossed simultaneously. Find the probability of getting
(i) exactly 1 head
(ii) at most 1 head
(iii) at least 1 head

Answer:

When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH and TT.
 Total number of possible outcomes = 4

(i) Let E be the event of getting exactly one head.

    Then, the favourable outcomes are HT and TH.
    ​Number of favourable outcomes = 2
  ∴ P(getting exactly 1 head)  = Number of favourable outcomesTotal number of possible outcomes =24  = 12

 (ii) Let E be the event of getting at most one head.
    Then, the favourable outcomes are HT, TH and TT.
    ​Number of favourable outcomes = 3
  ∴ P (getting at most 1 head)  = Number of favourable outcomesTotal number of possible outcomes =34  

(iii) Let E be the event of getting at least one head.
    Then, the favourable outcomes are HT, TH and HH
    ​Number of favourable outcomes = 3
  ∴ P (getting at least 1 head)  = Number of favourable outcomesTotal number of possible outcomes =34  



Page No 928:

Question 4:

A die is thrown once. Find the probability of getting
(i) an even number
(ii) a number less than 5
(iii) a number greater than 2
(iv) a number between 3 and 6
(v) a number other than 3
(vi) the number 5.

Answer:

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.
Total number of possible outcomes = 6

(i)
   Let E be the event of getting an even number.
   Then, the favourable outcomes are 2, 4 and 6.
   Number of favourable outcomes = 3
∴ Probability of getting an even number =  (E) = Number of favourable outcomesTotal number of possible outcomes = 36 = 12
(ii)
   Let E be the event of getting a number less than 5.
   Then, the favourable outcomes are 1, 2, 3, 4
   Number of favourable outcomes = 4
∴ Probability of getting a number less than 5 =  (E) = Number of favourable outcomesTotal number of possible outcomes = 46 = 23

 (iii)
   Let E be the event of getting a number greater than 2.     
   Then, the favourable outcomes are 3, 4, 5 and 6.
   Number of favourable outcomes = 4
∴ Probability of getting a number greater than 2 =  (E) = Number of favourable outcomesTotal number of possible outcomes = 46 = 23
(iv)
    Let E be the event of getting a number between 3 and 6.

   Then, the favourable outcomes are 4, 5   
   Number of favourable outcomes = 2
∴ Probability of getting a number between 3 and 6 =  (E) = Number of favourable outcomesTotal number of possible outcomes = 26 = 13
(v)
    Let E be the event of getting a number other than 3.  
   Then, the favourable outcomes are 1, 2, 4, 5 and 6.     
    Number of favourable outcomes = 5
∴ Probability of getting a number other than 3 =  (E) = Number of favourable outcomesTotal number of possible outcomes = 56 
(vi)
   Let E be the event of getting the number 5.  
    Then, the favourable outcome is 5.      
    Number of favourable outcomes = 1
∴ Probability of getting the number 5 =  (E) = Number of favourable outcomesTotal number of possible outcomes = 16 

Page No 928:

Question 5:

A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.

Answer:

Let E be the event of getting a consonant.

Out of 26 letters of English alphabets, there are 21 consonants.

∴ P(getting a consonant) = P(E) = Number of outcomes favourable to ENumber of all possible outcomes
                                                    = 2126

Thus, the probability of getting a consonant is 2126.

Page No 928:

Question 6:

A child has die whose 6 faces show the letters given below.
 

A B C A A B

The die is thrown once. What is the probability of getting (i) A, (ii) B?

Answer:


When the die is thrown once, then any one of the six faces can show up.

∴ Total number of outcomes = 6

(i) There are 3 faces on the die showing the letter A.

Favourable number of outcomes = 3

∴ P(Getting the letter A) = Favourable number of outcomesTotal number of outcomes=36=12

(ii) There are 2 faces on the die showing the letter B.

Favourable number of outcomes = 2

∴ P(Getting the letter B) = Favourable number of outcomesTotal number of outcomes=26=13

Page No 928:

Question 7:

12 defective pens are accidentally mixed up the 132 good ones. It is not possible to just look at a pen and tell whether it is defective or not. One pen is taken out at random from the lot. Find the probability that the pen taken out is (i) a good one, (ii) defective

Answer:


Number of defective pens in the lot = 12

Number of good pens in the lot = 132

Total number of pens in the lot = 12 + 132 = 144

∴ Total number of outcomes = 144

(i) There are 132 good pens in the lot. Out of these pens, one good pen can be taken out in 132 ways.

Favourable number of outcomes = 132

∴ P(Pen taken out is good one) = Favourable number of outcomesTotal number of outcomes=132144=1112

(ii) There are 12 defective pens in the lot. Out of these pens, one defective pen can be taken out in 12 ways.

Favourable number of outcomes = 12

∴ P(Pen taken out is defective) = Favourable number of outcomesTotal number of outcomes=12144=112

Page No 928:

Question 8:

In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?

Answer:

Total number of lottery tickets = 10 + 25 = 35
Number of prizes = 10

Let  E be the event of getting a prize.

∴ P(getting a prize) = P(E) = Number of outcomes favourable to ENumber of all possible outcomes
                                            = 1035=27

Thus, the probability of getting a prize is 27.

Page No 928:

Question 9:

250 lottery tickets were sold and there are 5 prizes on these tickets. If Kunal has purchased one lottery ticket, what is the probability that he wins a prize?

Answer:

Total number of tickets = 250
Kunal wins a prize if he gets a ticket that assures a prize.
Number of tickets on which prizes are assured = 5
     ∴ (Kunal wins a prize)  =  5250 = 150

Page No 928:

Question 10:

If the probability of winning a game is 0.7, what is the probability of losing it?

Answer:

For any event E, P(E) + P(not E) = 1

Let probability of winning a game = P(E) = 0.7

∴ P(winning a game) + P(losing a game) = 1
⇒ P(losing a game) = 1 − 0.7
                                = 0.3

Thus, the probability of losing a game is 0.3.

Page No 928:

Question 11:

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the numbers 1, 2, 3,..., 12 as shown in the figure. What is the probability that it will  point to

(i) 6?
(ii) an even number
(iii) a prime number?
(iv) a number which is a multiple of 5?
(v) a number which is a factor of 8?

Answer:


The arrow can come to rest at any one of the numbers 1, 2, 3, 4,..., 12.

∴ Total number of outcomes = 12

(i) There is only one 6 in the figure. So, there is only one way when the arrow will come to rest pointing to the number 6.

Favourable number of outcomes = 1 

∴ P(Arrow will point to 6) = Favourable number of outcomesTotal number of outcomes=112

(ii) There are six even numbers in the figure. These are 2, 4, 6, 8, 10 and 12. So, there are 6 ways when the arrow will come to rest pointing to an even number.

Favourable number of outcomes = 6

∴ P(Arrow will point to an even number) = Favourable number of outcomesTotal number of outcomes=612=12

(iii) There are five prime numbers in the figure. These are 2, 3, 5, 7 and 11. So, there are 5 ways when the arrow will come to rest pointing to a prime number.

Favourable number of outcomes = 5

∴ P(Arrow will point to a prime number) = Favourable number of outcomesTotal number of outcomes=512

(iv) There are two multiples of 5 in the figure. These are 5 and 10. So, there are 2 ways when the arrow will come to rest pointing to a number which is a multiple of 5.

Favourable number of outcomes = 2

∴ P(Arrow will point to a number which is a multiple of 5) = Favourable number of outcomesTotal number of outcomes=212=16

(v) There are four factors of 8 in the figure. These are 1, 2, 4 and 8. So, there are 4 ways when the arrow will come to rest pointing to a number which is a factor of 8.

Favourable number of outcomes = 4

∴ P(Arrow will point to a number which is a factor of 8) = Favourable number of outcomesTotal number of outcomes=412=13

Page No 928:

Question 12:

17 cards numbered 1, 2, 3, 4, .... ,17 are put in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the card drawn bears (i) an odd number (ii) a number divisible by 5.

Answer:

Total number of cards = 17

(i) Let  E1 be the event of choosing an odd number.

These numbers are 1, 3, 5, 7, 9, 11, 13, 15 and 17.

∴ P(getting an odd number) = P(E1) = Number of outcomes favourable to E1Number of all possible outcomes
                                                           = 917

Thus, the probability that the card drawn bears an odd number is 917.

(i) Let  E2 be the event of choosing a number divisible by 5.

These numbers are 5, 10 and 15.
∴ P(getting a number divisible by 5) = P(E2) = Number of outcomes favourable to E2Number of all possible outcomes
                                                                        = 317

Thus, the probability that the card drawn bears a number divisible by 5 is 317.



Page No 929:

Question 13:

A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls. A ball is drawn at random from the bag. Find the probability that it is

(i) black,
(ii) not green, 
(iii) red or white,
(iv) neither red nor green.

Answer:

Total number of balls = 15

(i) Number of black balls = 2

∴  P(getting a black ball) = Number of favourable outcomesNumber of all possible outcomes
                                         = 215

Thus, the probability of getting a black ball is 215.

(ii) Number of balls which are not green = 4 + 5 + 2 = 11

∴  P(getting a ball which is not green) = Number of favourable outcomesNumber of all possible outcomes
                                                             = 1115

Thus, the probability of getting a ball which is not green is 1115.

(iii) Number of balls which are either red or white = 4 + 5 = 9

∴  P(getting a ball which is red or white) = Number of favourable outcomesNumber of all possible outcomes
                                                                  = 915=35

Thus, the probability of getting a ball which is red or white is 35.

(iv) Number of balls which are neither red nor green = 4 + 2 = 6

∴  P(getting a ball which is neither red nor green) = Number of favourable outcomesNumber of all possible outcomes
                                                                                = 615=25

Thus, the probability of getting a ball which is neither red nor green is 25.

Page No 929:

Question 14:

A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is (i) white or blue, (ii) neither white nor black.

Answer:


Number of white balls in the bag = 5

Number of red balls in the bag = 7

Number of black balls in the bag = 4

Number of blue balls in the bag = 2

Total number of balls in the bag = 5 + 7 + 4 + 2 = 18

∴ Total number of outcomes = 18

(i) There are 7 balls (5 white and 2 blue) in the bag which are either white or blue. So, there are 7 ways to draw a ball from the bag which is white or blue.

Favourable number of outcomes = 7

∴ P(Drawn ball is white or blue) = Favourable number of outcomesTotal number of outcomes=718

(ii) There are 9 balls (7 red and 2 blue) in the bag which are neither white nor black. So, there are 9 ways to draw a ball from the bag which is neither white nor black.

Favourable number of outcomes = 9

∴ P(Drawn ball is neither white nor black) = Favourable number of outcomesTotal number of outcomes=918=12

Page No 929:

Question 15:

A jar contains 24 marbles. Some of these are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 23. Find the number of blue marbles in the jar.

Answer:

Total number of marbles = 24.
Let the number of blue marbles be x.
Then, the number of green marbles = 24 − x

∴ P(getting a green marble) = Number of favourable outcomesNumber of all possible outcomes
                                             = 24-x24

But, P(getting a green marble) = 23 (given)

24-x24=233(24-x)=4872-3x=483x=72-483x=24x=8

Thus, the number of blue marbles in the jar is 8.

Page No 929:

Question 16:

A jar contains 54 marbles, each of which some are blue, some are green and some are white. The probability of selecting a blue marble at random is 13 and the probability of selecting a green marble at random is 49. How many white marbles does the jar contain?

Answer:

Total number of marbles = 54.

It is given that, P(getting a blue marble) = 13 and P(getting a green marble) = 49

Let P(getting a white marble) be x.

Since, there are only 3 types of marbles in the jar, the sum of probabilities of all three marbles must be 1.

13+49+x=13+49+x=1x=1-79x=29
∴ P(getting a white marble) = 29  ...(1)

Let the number of white marbles be n.

Then, P(getting a white marble) = n54  ... (2)

From (1) and (2),
n54=29n=2×549n=12

Thus, there are 12 white marbles in the jar.

Page No 929:

Question 17:

The probability of selecting a red ball at random from the jar that contains only red, blue and orange balls is 14. The probability of selecting a blue ball at random from the same jar is 13. If the jar contains 10 orange balls, find the total number of balls in the jar.

Answer:

It is given that,

P(getting a red ball) = 14 and P(getting a blue ball) = 13

Let P(getting an orange ball) be x.

Since, there are only 3 types of balls in the jar, the sum of probabilities of all the three balls must be 1.

 14+13+x=1x=1-14-13x=12-3-412x=512\
∴ P(getting an orange ball) = 512

Let the total number of balls in the jar be n.

∴ P(getting an orange ball) = 10n
10n=512n=24

Thus, the total number of balls in the jar is 24.

Page No 929:

Question 18:

A bag contains 18 balls out of which x balls are red.

(i) If the ball is drawn at random from the bag, what is the probability that it is not red?

(ii) If two more red balls are put in the bag, the probability of drawing a red ball will be 98 times the probability of drawing a red ball in the first case. Find the value of x.

Answer:

Total number of balls = 18.
Number of red balls = x.

(i) Number of balls which are not red = 18 − x

∴ P(getting a ball which is not red) = Number of favourable outcomesNumber of all possible outcomes
                                                        = 18-x18

Thus, the probability of drawing a ball which is not red is 18-x18.

(ii) Now, total number of balls = 18 + 2 = 20.
Number of red balls now = x + 2.

P(getting a red ball now) = Number of favourable outcomesNumber of all possible outcomes
                                         = x+220

and P(getting a red ball in first case) = Number of favourable outcomesNumber of all possible outcomes
                                                           = x18

Since, it is given that probability of drawing a red ball now will be 98 times the probability of drawing a red ball in the first case.

Thus, x+220=98×x18
144(x+2)=180x144x+288=180x36x=288x=28836=8

Thus, the value of x is 8.

Page No 929:

Question 19:

A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball find the number of black balls in the bag.       [CBSE 2017]

Answer:

The number of white balls in a bag = 15.
Let the number of black balls in that bag be x.
Then, the total number of balls in bag = 15+x.

Now, P(black ball)=x15+x and P(white ball)=1515+x.
According to question, P(black ball) = 3 P(white ball)
x15+x=3×1515+xx15+x=4515+xx=45.
Hence, number of black balls in bag = x = 45.

Page No 929:

Question 20:

A box contains 90 discs, which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a number divisible by 5

Answer:

Ans

Page No 929:

Question 21:

A bag contains some balls of which x are white, 2x are black are 3x are red. A ball is selected at random. What is the probability that it is (i) not red? (ii) white?

Answer:


Number of white balls in the bag = x

Number of black balls in the bag = 2x

Number of red balls in the bag = 3x

Total number of balls in the bag = x + 2x + 3x = 6x

∴ Total number of outcomes = 6x

(i) There are 3x non-red balls (x white balls and 2x black balls) in the bag. So, there are 3x ways to draw a ball from the bag which is not red.

Favourable number of outcomes = 3x

∴ P(Selected ball is not red) = Favourable number of outcomesTotal number of outcomes=3x6x=12

(ii) There are x white balls in the bag. So, there are x ways to draw a ball from the bag which is white.

Favourable number of outcomes = x

∴ P(Selected ball is white) = Favourable number of outcomesTotal number of outcomes=x6x=16



Page No 930:

Question 22:

Cards, marked with numbers 5 to 50, are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is (i) a prime number less than 10 (ii) a perfect square.

Answer:

All possible outcomes are 5, 6, 7, 8...................50.

Number of all possible outcomes = 46

(i)  Out of the given numbers, the prime numbers less than 10 are 5 and 7.
    Let E1 be the event of getting a prime number less than 10.
  Then, number of favourable outcomes =  2​                         
   ∴ P (getting a prime number less than 10) =  246 = 123   

(ii) Out of the given numbers, the perfect squares are 9, 16, 25, 36 and 49.  
   Let E2 be the event of getting a perfect square.
   Then, number of favourable outcomes = 5​                         
   ∴ P (getting a perfect square) =  546      

Page No 930:

Question 23:

A box contains cords numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (i) a prime number, (ii) a composite number, (iii) a number divisible by 3.

Answer:


There are 20 cards in the box. One card can drawn at random from the box in 20 ways.

∴ Total number of outcomes = 20

(i) The prime number cards in the box are 2, 3, 5, 7, 11, 13, 17 and 19. There are 8 prime number cards in the box. So, there are 8 ways to draw a card from the box which is a prime number card.

Favourable number of outcomes = 8

∴ P(Number on the drawn card is a prime number) = Favourable number of outcomesTotal number of outcomes=820=25

(ii) The composite number cards in the box are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 and 20. There are 11 composite number cards in the box. So, there are 11 ways to draw a card from the box which is a composite number card.

Favourable number of outcomes = 11

∴ P(Number on the drawn card is a composite number) = Favourable number of outcomesTotal number of outcomes=1120

(iii) The cards with number divisible by 3 on them in the box are 3, 6, 9, 12, 15 and 18. There are 6 cards with number divisible by 3 on them in the box. So, there are 6 ways to draw a card from the box with number divisible by 3 on them.

Favourable number of outcomes = 6

∴ P(Number on the drawn card is a number divisible by 3) = Favourable number of outcomesTotal number of outcomes=620=310

Page No 930:

Question 24:

A box contains cards bearing numbers 6 to 70. If one card is drawn at random from the box, find the probability that it bears

(i) a one-digit number,
(ii) a number divisible by 5,
(iii) an odd number less than 30,
(iv) a composite number between 50 and 70.

Answer:

​Given number 6, 7, 8, .... , 70 form an AP with a = 6 and d = 1.
Let Tn = 70. Then,
6 + (n − 1)1 = 70
⇒ 6 + n  − 1 = 70
⇒ n = 65

Thus, total number of outcomes = 65.

(i) Let E1 be the event of getting a one-digit number.

Out of these numbers, one-digit numbers are 6, 7, 8 and 9.

Number of favourable outcomes = 4.

∴ P(getting a one-digit number) = P(E1) = Number of outcomes favourable to E1Number of all possible outcomes
                                                                 = 465

Thus, the probability that the card bears a one-digit number is 465.


(ii) Let E2 be the event of getting a number divisible by 5.

Out of these numbers, numbers divisible by 5 are 10, 15, 20, ... , 70.
Given number 10, 15, 20, .... , 70 form an AP with a = 10 and d = 5.
Let Tn = 70. Then,
10 + (n − 1)5 = 70
⇒ 10 + 5n  − 5 = 70
⇒ 5n = 65
n = 13

Thus, number of favourable outcomes = 13.

∴ P(getting a number divisible by 5) = P(E2) = Number of outcomes favourable to E2Number of all possible outcomes
                                                                        = 1365=15

Thus, the probability that the card bears a number divisible by 5 is 15.

(iii) Let E3 be the event of getting an odd number less than 30.

Out of these numbers, odd numbers less than 30 are 7, 9, 11, ... , 29.
Given number 7, 9, 11, .... , 29 form an AP with a = 7 and d = 2.
Let Tn = 29. Then,
7 + (n − 1)2 = 29
⇒ 7 + 2n  − 2 = 29
⇒ 2n = 24
⇒ n = 12

Thus, number of favourable outcomes = 12.

∴ P(getting an odd number less than 30) = P(E3) = Number of outcomes favourable to E3Number of all possible outcomes
                                                                              = 1265

Thus, the probability that the card bears an odd number less than 30 is 1265.

(iv) Let E4 be the event of getting a composite number between 50 and 70.

Out of these numbers, composite numbers between 50 and 70 are 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68 and 69.

Number of favourable outcomes = 15.

∴ P(getting a composite number between 50 and 70) = P(E4) = Number of outcomes favourable to E4Number of all possible outcomes
                                                                                                  = 1565=313

Thus, the probability that the card bears a composite number between 50 and 70 is 313.

Page No 930:

Question 25:

Cards marked with numbers 1, 3, 5, ..., 101 are placed in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) less than 19,
(ii) a prime number less than 20.

Answer:

​Given number 1, 3, 5, ..., 101 form an AP with a = 1 and d = 2.
Let Tn = 101. Then,
1 + (n − 1)2 = 101
⇒ 1 + 2n  − 2 = 101
⇒ 2n = 102
n = 51

Thus, total number of outcomes = 51.

(i) Let E1 be the event of getting a number less than 19.

Out of these numbers, numbers less than 19 are 1, 3, 5, ... , 17.
Given number 1, 3, 5, .... , 17 form an AP with a = 1 and d = 2.
Let Tn = 17. Then,
1 + (n − 1)2 = 17
⇒ 1 + 2n  − 2 = 17
⇒ 2n = 18
⇒ n = 9

Thus, number of favourable outcomes = 9.

∴ P(getting a number less than 19) = P(E1) = Number of outcomes favourable to E1Number of all possible outcomes
                                                                     = 951=317

Thus, the probability that the number on the drawn card is less than 19 is 317.

(ii) Let E2 be the event of getting a prime number less than 20.

Out of these numbers, prime numbers less than 20 are 3, 5, 7, 11, 13, 17 and 19.

Thus, number of favourable outcomes = 7.

∴ P(getting a prime number less than 20) = P(E2) = Number of outcomes favourable to E2Number of all possible outcomes
                                                                                = 751

Thus, the probability that the number on the drawn card is a prime number less than 20 is 751.

Page No 930:

Question 26:

Cards bearing numbers 1, 3, 5, .... , 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing

(i) a prime number less than 15,
(ii) a number divisible by 3 and 5.

Answer:

Given number 1, 3, 5, .... , 35 form an AP with a = 1 and d = 2.
Let Tn = 35. Then,
1 + (n − 1)2 = 35
⇒ 1 + 2n  − 2 = 35
⇒ 2n = 36
n = 18

Thus, total number of outcomes = 18.

(i) Let E1 be the event of getting a prime number less than 15.

Out of these numbers, prime numbers less than 15 are 3, 5, 7, 11 and 13.

Number of favourable outcomes = 5.

∴ P(getting a prime number less than 15) = P(E1) = Number of outcomes favourable to E1Number of all possible outcomes
                                                                                = 518

Thus, the probability of getting a card bearing a prime number less than 15 is 518.


(ii) Let E2 be the event of getting a number divisible by 3 and 5.

Out of these numbers, number divisible by 3 and 5 means number divisible by 15 is 15.

Number of favourable outcomes = 1.

∴ P(getting a number divisible by 3 and 5) = P(E2) = Number of outcomes favourable to E2Number of all possible outcomes
                                                                                  = 118

Thus, the probability of getting a card bearing a number divisible by 3 and 5 is 118.

Page No 930:

Question 27:

A box contains cards numbered 3, 5, 7, 9, .... , 35, 37. A card is drawn at random from the box. Find the probability that the number on the card is a prime number.

Answer:

Given numbers 3, 5, 7, 9, .... , 35, 37 form an AP with a = 3 and d = 2.
Let Tn = 37. Then,
3 + (n − 1)2 = 37
⇒ 3 + 2n − 2 = 37
⇒ 2n = 36
n = 18

Thus, total number of outcomes = 18.

​Let E be the event of getting a prime number.

Out of these numbers, the prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37.

Number of favourable outcomes = 11.

∴ P(getting a prime number) = P(E) = Number of outcomes favourable to ENumber of all possible outcomes
                                                          = 1118

Thus, the probability that the number on the card is a prime number is 1118.

Page No 930:

Question 28:

Cards numbered 1 to 30 are put in a bag. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) not divisible by 3,
(ii) a prime number greater than 7,
(iii) not a perfect square number.

Answer:

Total number of outcomes = 30.

(i) ​Let E1 be the event of getting a number not divisible by 3.

Out of these numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30.

Number of favourable outcomes = 30 − 10 = 20

∴ P(getting a number not divisible by 3) = P(E1) = Number of outcomes favourable to E1Number of all possible outcomes
                                                                              = 2030=23

Thus, the probability that the number on the card is not divisible by 3 is 23.

(ii) ​Let E2 be the event of getting a prime number greater than 7.

Out of these numbers, prime numbers greater than 7 are 11, 13, 17, 19, 23 and 29.

Number of favourable outcomes = 6

∴ P(getting a prime number greater than 7) = P(E2) = Number of outcomes favourable to E2Number of all possible outcomes
                                                                                   = 630=15

Thus, the probability that the number on the card is a prime number greater than 7 is 15.

(iii) ​Let E3 be the event of getting a number which is not a perfect square number.

Out of these numbers, perfect square numbers are 1, 4, 9, 16 and 25.

Number of favourable outcomes = 30 − 5 = 25

∴ P(getting non-perfect square number) = P(E3) = Number of outcomes favourable to E3Number of all possible outcomes
                                                                              = 2530=56

Thus, the probability that the number on the card is not a perfect square number is 56.

Page No 930:

Question 29:

A box contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) divisible by 2 or 3,
(ii) a prime number.

Answer:

Total number of outcomes = 25

(i) Let E1 be the event of getting a card divisible by 2 or 3.

Out of the given numbers, numbers divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24.
Out of the given numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21 and 24.
Out of the given numbers, numbers divisible by both 2 and 3 are 6, 12, 18 and 24.

Number of favourable outcomes = 16

∴ P(getting a card divisible by 2 or 3) = P(E1) = Number of outcomes favourable to E1Number of all possible outcomes
                                                                           = 1625

Thus, the probability that the number on the drawn card is divisible by 2 or 3 is 1625.

(ii) Let E2 be the event of getting a prime number.

Out of the given numbers, prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23.

Number of favourable outcomes = 9

∴ P(getting a prime number) = P(E2) = Number of outcomes favourable to E2Number of all possible outcomes
                                                            = 925

Thus, the probability that the number on the drawn card is a prime number is 925.

Page No 930:

Question 30:

Tickets numbered 2, 3, 4, 5, ..., 100, 101 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is
(i) an even number
(ii) a number less than 16
(iii) a number which is a perfect square
(iv) a prime number less than 40

Answer:

All possible outcomes are  2, 3, 4, 5................101.
Number of all possible outcomes = 100

(i) Out of these the numbers that are even = 2, 4, 6, 8...................100
     Let E1 be the event of getting an even number.
    Then, number of favourable outcomes = 50​            Tn= 100     2 + (n -1) ×2 = 100, n = 50 
   ∴ P (getting an even number) = 50100  = 12

(ii) Out of these, the numbers that are less than 16 = 2,3,4,5,..................15.
     Let E2 be the event of getting a number less than 16.
  Then,​ number of favourable outcomes =​ 14
    ∴ P (getting a number less than 16) = 14100  = 750

(iii) Out of these, the numbers that are perfect squares =  4, 9,16,25, 36, 49, 64, 81 and 100 
      Let E3 be the event of getting a number that is a perfect square.

  Then,​ number of favourable outcomes = 9
 P (getting a number that is a perfect square) = 9100 

(iv) Out of these, prime numbers less than 40 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
      Let E4 be the event of getting a prime number less than 40.
    Then,​ number of favourable outcomes =​ 12
    ∴ P (getting a prime number less than 40) = 12100  = 325



Page No 931:

Question 31:

In a family of 3 children, find the probability of having at least one boy.

Answer:

All possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB and GGG.

Number of all possible outcomes = 8

Let E be the event of having at least one boy.

Then the outcomes are BBB, BBG, BGB, GBB, BGG, GBG and GGB.

Number of possible outcomes = 7
∴ P(Having at least one boy) = P(E) = Number of outcomes favourable to ENumber of all possible outcomes
                                                           = 78

Thus, the probability of having at least one boy is 78.

Page No 931:

Question 32:

Two different dice are thrown together. Find the probability that the numbers obtained have
(i) even sum
(ii) even product.

Answer:

Total number of possible outcomes is 36.
(i) The favorable outcomes are (1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6).
P(the sum is even)=1836=12.

(ii) The favorable outcomes are (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).
P(the product is even)=2736=34.

Page No 931:

Question 33:

Two different dice are thrown together. Find the probability that the numbers obtained
(i) have a sum less than 7
(ii) have a product less than 16
(iii) is a doublet of odd numbers.          [CBSE 2017]

Answer:

Total number of possible outcomes is 36.
(i) The favorable outcomes for sum of numbers on dices less than 7 are (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).
P(the sum of numbers appeared is less than 7)=1536=512.

(ii) The favorable outcomes for product of numbers on dices less than 16 are (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), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1), (6,2).
P(the product of numbers appeared is less than 16)=2536.

(iii) The favorable outcomes are (1,1), (3,3), (5,5).
P(the doublet of odd numbers)=336=112.

Page No 931:

Question 34:

Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25?       [CBSE 2017]

Answer:

Total number of possible outcomes for Peter = 36.
Possible outcomes for Peter to get product 25 is (5,5).
Total number of possible outcomes for Rina = 6.
Possible outcomes for Rina to get the number whose square is 25 is 5.

Now, P(Peter will get 25)=136.
And, P(Rina will get 25)=16.
Since 136<16, So Rina has better chances of getting 25.

Page No 931:

Question 35:

Find the probability of getting the sum of two numbers, less than 3 or more than 11, when a pair of distinct dice is thrown together.     [CBSE 2017]

Answer:

Favorable outcomes for sum of numbers on dice less than 3 or more than 11 are (1,1), (6,6).
Required probability = P(sum is less than 3 or more than 11) = 236=118.

Page No 931:

Question 36:

The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?    [CBSE 2017]

Answer:

Total number of apples = 900.
P(a rotten apple) = 0.18
number of rotten applestotal number of apples=0.18number of rotten apples900=0.18The number of rotten apples=0.18×900=162 apples.

Page No 931:

Question 37:

What is the probability that an ordinary year has 53 Mondays?

Answer:

An ordinary year has 365 days consisting of 52 weeks and 1 day.

This day can be any day of the week.

∴ P(of this day to be Monday) = 17

Thus, the probability that an ordinary year has 53 Mondays is 17.

Page No 931:

Question 38:

Find the probability that a leap year selected at random will contain 53 Sundays.

Answer:

A leap year has 366 days with 52 weeks and 2 days.

Now, 52 weeks conatins 52 sundays.

The remaining two days can be:
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday

Out of these 7 cases, there are two cases favouring it to be Sunday.

∴ P(a leap year having 53 Sundays) = Number of favourable outcomesNumber of all possible outcomes
                                                          = 27

Thus, the probability that a leap year selected at random will contain 53 Sundays is 27.

Page No 931:

Question 39:

A letter is chosen at random from the letters of the word ASSOCIATION. Find the probability that the chosen letter is a
(i) vowel
(ii) consonant
(iii) an S.

Answer:

Total numbers of letters in the given word ASSOCIATION = 11
(i) Number of vowels ( A, O, I, A, I, O) in the given word = 6
  ∴ P (getting a vowel) = 611

(ii) Number of consonants in the given word ( S, S, C, T, N) = 5
   ∴ P (getting a consonant) = 511

(iii) Number of S in the given word = 2
   ∴ P (getting an S) = 211

Page No 931:

Question 40:

A game consists of tossing a one-rupee coin three times, and noting its outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.

Answer:


When a coin is tossed three times, the possible outcomes are

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Total number of outcomes = 8

A game can be lost if the if all the tosses do not give the same result. The outcomes where tosses do not give the same result are

HHT, HTH, THH, HTT, THT, TTH

Favourable number of outcomes for losing the game = 6

∴ P(Game is lost) = Favourable number of outcomes for losing the gameTotal number of outcomes=68=34

Thus, the probability of losing the game is 34.

Page No 931:

Question 41:

The king, the jack and the 10 of spades are lost from a pack of 52 cards and a card is drawn from the remaining cards after shuffling. Find the probability of getting a
(i) red card
(ii) black jack
(iii) red king
(iv) 10 of hearts.          [CBSE 2017]

Answer:

Total number of cards in a deck is 52.
The number of cards left after loosing the King, the Jack and the 10 of spade = 52-3 = 49.
(i) Red card left after loosing King, the Jack and the 10 of spade will be 13 hearts + 13 diamond = 26 cards
P(red card)=2649.
(ii) Black jack left after loosing King, the Jack and the 10 of spade will be one only of club.
P(black jack)=149.
(iii) Red king left after loosing King, the Jack and the 10 of spade will be red of diamond and red of hearts i.e two cards.
P(red king)=249.
(iv) 10 of hearts will be one only after loosing King, the Jack and the 10 of spade
P(10 of hearts)=149.

Page No 931:

Question 42:

All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is

(i) a red card,
(ii) a face card,
(iii) a card of clubs.

Answer:

There are 6 red face cards. These are removed.

Thus, remaining number of cards = 52 − 6 = 46.

(i) Number of red cards now = 26 − 6 = 20.

∴ P(getting a red card) = Number of favourable outcomesNumber of all possible outcomes
                                     = 2046=1023

Thus, the probability that the drawn card is a red card is 1023.

(ii) Number of face cards now = 12 − 6 = 6.

∴ P(getting a face card) = Number of favourable outcomesNumber of all possible outcomes
                                      = 646=323

Thus, the probability that the drawn card is a face card is 323.

(iii) Number of card of clubs = 12.

∴ P(getting a card of clubs) = Number of favourable outcomesNumber of all possible outcomes
                                             = 1246=623

Thus, the probability that the drawn card is a card of clubs is 623.



Page No 932:

Question 43:

All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is drawn from it. Find the probability that the drawn card is

(i) a black face card,
(ii) a red card.

Answer:

There are 4 kings, 4 queens and 4 aces. These are removed.

Thus, remaining number of cards = 52 − 4 − 4 − 4 = 40.

(i) Number of black face cards now = 2 (only black jacks).

∴ P(getting a black face card) = Number of favourable outcomesNumber of all possible outcomes
                                                = 240=120

Thus, the probability that the drawn card is a black face card is 120.

(ii) Number of red cards now = 26 − 6 = 20.

∴ P(getting a red card) = Number of favourable outcomesNumber of all possible outcomes
                                     = 2040=12

Thus, the probability that the drawn card is a red card is 12.

Page No 932:

Question 44:

A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the card drawn is neither a red card nor a queen.

Answer:

 Total number of all possible outcomes= 52
 There are 26 red cards (including 2 queens) and apart from these, there are 2 more queens.
 Number of cards, each one of which is either a red card or a queen = 26 + 2 = 28
 Let E be the event that the card drawn is neither a red card nor a queen.
  Then, the number of favourable outcomes =  (52 − 28) = 24 
  ∴ P( getting neither a red card nor a queen) = P (E) =   2452 = 613

Page No 932:

Question 45:

Five cards − the ten, jack, queen, king and ace of diamonds are well shuffled with their faces downwards. One card is then picked up at random.

(a) What is the probability that the drawn card is the queeen?
(b) If the queen is drawn and put aside and a second card is drawn, find the probability that the second card is (i) an ace, (ii) a queen.

Answer:

Total number of cards = 5.

(a) Number of queens = 1.

∴ P(getting a queen) = Number of favourable outcomesNumber of all possible outcomes
                                  = 15

Thus, the probability that the drawn card is the queen is 15.

(b) When the queen is put aside, number of remaining cards = 4.

(i) Number of aces = 1.

∴ P(getting an ace) = Number of favourable outcomesNumber of all possible outcomes
                                = 14

Thus, the probability that the drawn card is an ace is 14.

(ii) Number of queens = 0.

∴ P(getting a queen now) = Number of favourable outcomesNumber of all possible outcomes
                                          = 04=0

Thus, the probability that the drawn card is a queen is 0.

Page No 932:

Question 46:

A card is drawn at random form a well-shuffled deck of playing cards. Find the probability that the card drawn is
(i) a card of spades of an ace
(ii) a red king
(iii) either a king or a queen
(iv) neither a king nor a queen.

Answer:

Total number of all possible outcomes= 52
(i) Number of spade cards = 13
   Number of aces = 4 (including 1 of spade)

   Therefore, number of spade cards and aces = (13 + 4 − 1) = 16  
   ∴ P( getting a spade or an ace card) = 1652 = 413

(ii) Number of  red kings = 2
     ∴ P( getting a red king) =  252 = 126

(iii) Total number of kings = 4
     Total number of queens = 4
      Let E be the event of getting either a king or a queen.
      Then, the favourable outcomes =  4 + 4 = 8 
    ∴ P( getting a king or a queen) = P (E) =   852 = 213

(iv)  Let E be the event of getting either a king or a queen. Then, ( not E) is the event that drawn card is neither a king nor a
       queen.     
      Then, P(getting a king or a queen ) =  213 
     Now, ​P (E) + ​P (not E) = 1
     ∴ ​P(getting neither a king nor a queen ) = 1-213 = 1113

Page No 932:

Question 47:

Find the probability that a number selected at random from the numbers 3, 4, 4,4 5, 5, 6, 6, 6, 7 will be their mean.

Answer:


The number are 3, 4, 4, 4, 5, 5, 6, 6, 6, 7. One number can be selected at random from the given 10 numbers in 10 ways.

Total number of outcomes = 10

Mean of the given numbers = 3+4+4+4+5+5+6+6+6+710=5010=5

Now, there are two 5's among the given numbers. So, there are 2 ways to select the number 5 i.e. the mean of the given numbers.

Favourable number of outcomes = 2

∴ P(Selected number is the mean of the given number) = P(Selecting the number 5) = Favourable number of outcomesTotal number of outcomes=210=15

Thus, the probability that a number selected at random from the given numbers will be their mean is 15.



Page No 944:

Question 1:

If P(E) denotes the probability of an event E then               [CBSE 2013C]

(a) P(E) < 0
(b) P(E) > 1
(c) 0 ≤ P(E) ≤ 1
(d) −1 ≤ P(E) ≤ 1

Answer:

Since, number of elements in the set of favourable cases is less than or equal to the number of elements in the set of whole number of cases,
their ratio always end up being 1 or less than 1.
Also, their ratio can never be negative.

Thus, probability of an event always lies between 0 and 1.
i.e. 0 ≤ P(E) ≤ 1

Hence, the correct answer is option (c).

Page No 944:

Question 2:

If the probability of occurence of an event is p then the probability of non-happening of this event is      [CBSE 2013C]

(a) (p − 1)
(b) (1 − p)
(c) p
(d) 1-1p

Answer:

P(occurence of an event) = p
P(non-occurence of this event) = 1 − p

Hence, the correct answer is option (b).

Page No 944:

Question 3:

What is the probability of an impossible event?
(a) 12
(b) 0
(c) 1
(d) none of these

Answer:

(b) 0

The probability of an impossible event is 0.

Page No 944:

Question 4:

What is the probability of a sure event?
(a) 0
(b) 12
(c) 1
(d) none of these

Answer:

(c) 1

The probability of a sure event is 1.

Page No 944:

Question 5:

Which of the following cannot be the probability of an event?       [CBSE 2013C]      

(a) 1.5
(b) 35
(c) 25%
(d) 0.3

Answer:

The probability of an event cannot be greater than 1.
Thus, the probability of an event cannot be 1.5.

Hence, the correct answer is option (a).

Page No 944:

Question 6:

A number is selected at random from the numbers 1 to 30. What is the probability that the selected number is a prime number?                                                                                                                 [CBSE 2014]

(a) 23
(b) 16
(c) 13
(d) 1130

Answer:

Total number of outcomes = 30.

Prime numbers in 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

Number of favourable outcomes = 10.

∴ P(getting a prime number) = Number of favourable outcomesNumber of all possible outcomes
                                               = 1030=13

Thus, the probability that the selected number is a prime number is 13.

Hence, the correct answer is option (c).

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Question 7:

The probability that a number selected at random from the numbers 1, 2, 3, ..., 15 is a multiple of 4, is  [CBSE 2014]

(a) 415
(b) 215
(c) 15
(d) 13

Answer:

Total number of outcomes = 15.

Out of the given numbers, multiples of 4 are 4, 8 and 12.

Numbers of favourable outcomes = 3.

∴ P(getting a multiple of 4) = Number of favourable outcomesNumber of all possible outcomes
                                             = 315=15

Thus, the probability that a number selected is a multiple of 4 is 15.

Hence, the correct answer is option (c).

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Question 8:

A box conatins cards numbered 6 to 50. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square is                           [CBSE 2013]

(a) 145
(b) 215
(c) 445
(d) 19

Answer:

Total number of outcomes = 45.

Out of the given numbers, perfect square numbers are 9, 16, 25, 36 and 49.

Numbers of favourable outcomes = 5.

∴ P(getting a number which is a perfect square) = Number of favourable outcomesNumber of all possible outcomes
                                                                             = 545=19

Thus, the probability that the drawn card has a number which is a perfect square is 19.

Hence, the correct answer is option (d).



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Question 9:

A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears prime number less than 23 is                                                           [CBSE 2013]
    
(a) 790
(b) 19
(c) 415
(d) 889

Answer:

Total number of discs = 90.

Out of the given numbers, prime numbers less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19.

Numbers of favourable outcomes = 8.

∴ P(getting a prime number which is less than 23) = Number of favourable outcomesNumber of all possible outcomes
                                                                                 = 890=445

Thus, the probability that the disc bears prime number less than 23 is 445.

Disclaimer: There is a misprinting in the question. If they ask for the probability of all prime numbers less than 90, then we get 415.

Hence, the correct answer is option (c).

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Question 10:

Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is                                   [CBSE 2012]

(a) 12
(b) 25
(c) 310
(d) 59

Answer:

​Total number of cards = 10.

Out of the given numbers, prime numbers are 2, 3, 5, 7 and 11.

Numbers of favourable outcomes = 5.

∴ P(getting a prime number) = Number of favourable outcomesNumber of all possible outcomes
                                               = 510=12

Thus, the probability of getting a card with a prime number is 12.

Hence, the correct answer is option (a).

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Question 11:

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number, which is a multiple of 7, is                                                  [CBSE 2013C]

(a) 17
(b) 18
(c) 15
(d) 740

Answer:

​Total number of tickets = 40.

Out of the given numbers, multiples of 7 are 7, 14, 21, 28 and 35.

Numbers of favourable outcomes = 5.

∴ P(getting a number which is a multiple of 7) = Number of favourable outcomesNumber of all possible outcomes
                                                                           = 540=18

Thus, the probability that the selected ticket has a number, which is a multiple of 7, is 18.

Hence, the correct answer is option (b).

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Question 12:

Which of the following cannot be the probability of an event?
(a) 13
(b) 0.3
(c) 33%
(d) 76

Answer:

(d) 76

Explanation:
Probability of an event can't be more than 1.

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Question 13:

If the probability of winning a game is 0.4, the probability of losing it is
(a) 0.96
(b) 10.4
(c) 0.6
(d) none of these

Answer:

(c) 0.6

Explanation:
Let E be the event of winning a game.
Then, ( not E) is the event of not winning the game or of losing the game.
Then, P(E) = 0.4
Now, P(E) + ​P(not E) = 1 
⇒ 0.4 + ​P(not E) = 1
​⇒​ P(not E) = 1− 0.4 = 0.6
∴ P(losing the game) = ​P(not E) ​= 0.6

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Question 14:

If an event cannot occur then its probability is
(a) 1
(b) 12
(c) 34
(d) 0

Answer:

(d) 0

If an event cannot occur, its probability is 0.

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Question 15:

There are 20 tickets numbered as 1, 2, 3,..., 20 respectively. One ticket is drawn at random. What is the probability that the number on the ticket drawn is a multiple of 5?
(a) 14
(b) 15
(c) 25
(d) 310

Answer:

(b) 15

Explanation:
​Total number of tickets = 20
Out of the given ticket numbers, multiples of 5 are 5, 10, 15 and 20.
Number of favourable outcomes = 4
∴ P(getting a multiple of 5) =  420 = 15

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Question 16:

There are 25 tickets numbered 1, 2, 3, 4,..., 25 respectively. One ticket is draw at random. What is the probability that the number on the ticket is a multiple of 3 or 5?
(a) 25
(b) 1125
(c) 1225
(d) 1325

Answer:

(c)  1225 

Explanation:
​Total number of tickets = 25
Let E be the event of getting a multiple of 3 or 5.
Then,
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21 and 24.
Multiples of 5 are 5, 10, 15, 20 and 25.

Number of favourable outcomes = ( 8 + 5 − 1) = 12                     
∴ (getting a multiple of 3 or 5 )  = P (E) =  1225  

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Question 17:

Cards, each marked with one of the numbers 6, 7, 8,..., 15, are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with a number less than 10?
(a) 35
(b) 13
(c) 12
(d) 25

Answer:

(d)  25

Explanation:
All possible outcomes are 6, 7, 8................15.
Number of all possible outcomes = 10

Out of these, the numbers that are less than 10 are 6, 7, 8 and 9.  
Number of favourable outcomes = 4
 P(getting a number that is less than 10) = 410 = 25



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Question 18:

A die is thrown once. The probability of getting an even number is                            [CBSE 2013]

(a) 12
(b) 13
(c) 16
(d) 56

Answer:

​Total number of outcomes = 6.

Out of the given numbers, even numbers are 2, 4 and 6.

Numbers of favourable outcomes = 3.

∴ P(getting an even number) = Number of favourable outcomesNumber of all possible outcomes
                                               = 36=12

Thus, the probability of getting an even number is 12.

Hence, the correct answer is option (a).

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Question 19:

The probability of throwing a number greater than 2 with a fair die is                   [CBSE 2011]

(a) 25
(b) 56
(c) 13
(d) 23

Answer:

​Total number of outcomes = 6.

Out of the given numbers, numbers greater than 2 are 3, 4, 5 and 6.

Numbers of favourable outcomes = 4.

∴ P(getting a number greater than 2) = Number of favourable outcomesNumber of all possible outcomes
                                                           = 46=23

Thus, the probability of getting a number greater than 2 is 23.

Hence, the correct answer is option (d).

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Question 20:

A die is thrown once. The probability of getting an odd number greater than 3 is                     [CBSE 2013C]


(a) 13
(b) 16
(c) 12
(d) 0

Answer:

​Total number of outcomes = 6.

Out of the given numbers, odd number greater than 3 is 5.

Numbers of favourable outcomes = 1.

∴ P(getting an odd number greater than 3) = Number of favourable outcomesNumber of all possible outcomes
                                                                    = 16

Thus, the probability of getting an odd number greater than 3 is 16.

Hence, the correct answer is option (b).

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Question 21:

A die is  thrown once. The probability of getting a prime number is 
(a) 23 
(b) 13 
(c) 12
(d) 16

Answer:

(c) 12

Explanation:
  In a single throw of a die, the possible outcomes are:
    1, 2, 3, 4, 5, 6
   Total number of possible outcomes = 6
   Let E be the event of getting a prime number.
   Then, the favourable outcomes are 2, 3 and 5.                                        
   Number of favourable outcomes = 3
∴ Probability of getting a prime number =  (E) = Number of favorable outcomesTotal number of possible outcomes = 36 = 12

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Question 22:

Two dice are thrown together. The probability of getting the same number on the both dice is        [CBSE 2012]

(a) 12
(b) 13
(c) 16
(d) 112

Answer:

​Total number of outcomes = 36.

Getting the same number on both the dice means (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

Numbers of favourable outcomes = 6.

∴ P(getting the same number on both dice) = Number of favourable outcomesNumber of all possible outcomes
                                                                     = 636=16

Thus, the probability of getting the same number on both dice is 16.

Hence, the correct answer is option (c).

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Question 23:

The probability of getting 2 heads, when two coins are tossed, is                     [CBSE 2012]

(a) 1
(b) 34
(c) 12
(d) 14

Answer:

All possible outcomes are HH, HT, TH and TT.

​Total number of outcomes = 4.

Getting 2 heads means getting HH.

Numbers of favourable outcomes = 1.

∴ P(getting 2 heads) = Number of favourable outcomesNumber of all possible outcomes
                                 = 14

Thus, the probability of getting 2 heads is 14.

Hence, the correct answer is option (d).

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Question 24:

Two dice are thrown together. The probability of getting a doublet is

(a) 13

(b) 16

(c) 14

(d) 23

Answer:

(2) 16

Explanation: 
When two dice are thrown simultaneously, all possible outcomes are:
(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)
Number of all possible outcomes  = 36

 Let E be the event of getting a doublet.
 Then the favourable outcomes are:
  (1,1), (2,2) , (3,3) (4,4), (5,5), (6,6)
    Number of favourable outcomes = 6
∴ P(getting a doublet) = P ( E) =  636 = 16

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Question 25:

Two coins are tossed simultaneously. What is the probability of getting at most one head?
(a) 14
(b) 12
(c) 23
(d) 34

Answer:

(d) 34
 
Explanation:
 When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT.
  Total number of possible outcomes = 4

    Let E be the event of getting at most one head.

    Then, the favourable outcomes are HT, TH and TT.​
    ​Number of favourable outcomes = 3
  ∴ P(getting at most 1head) =   ​Number of favourable outcomesTotal number of possible outcomes =34  

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Question 26:

Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
(i) 12
(ii) 14
(iii) 38
(iv) 34

Answer:

(c) 38

 Explanation:
When 3 coins are tossed simultaneously, the possible outcomes are HHH, HHT, HTH, THH, THT, HTT, TTH and TTT.
   Total number of possible outcomes = 8

    Let E be the event of getting exactly two heads.

    Then, the favourable outcomes are  HHT, THH, and HTH.
    ​Number of favourable outcomes = 3
  ∴ Probability of getting exactly 2 heads =  P(E) = Number of favourable outcomesTotal number of possible outcomes =38  



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Question 27:

In a lottery, there are 8 prizes and 16 blanks. What is the probability of getting a prize?
(a) 12
(b) 13
(c) 23
(d) none of these

Answer:

(b)   13

Explanation:
Number of prizes = 8
Number of blanks = 16
Total number of tickets =  8 +16= 24
     ∴ P(getting a prize )  =  824 = 13

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Question 28:

In a lottery, there are 6 prizes and 24 blanks. What is the probability of not getting a prize?
(a) 34
(b) 35
(c) 45
(d) none of these

Answer:

(c)  45

Explanation:
Number of prizes = 6
Number of blanks = 24
Total number of tickets =  6 + 24= 30
     ∴ P(not getting a prize )  =  2430 = 45

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Question 29:

A box contains 3 blue, 2 white and 4 red marbles. If a marbles. If a marble is drawn at random from the box, what is the probability that it will not be a white marble?
(a) 13
(b) 49
(c) 79
(d) 29

Answer:

(c) 79

Explanation:
Total possible outcomes = Total number of marbles

                                       = ( 3 + 2 + 4) = 9 
Let E be the event of not getting a white marble.
It means the marble can be either blue or red but not white.
Number of favourable outcomes =  (3 + 4) = 7 marbles
   ∴ P(not getting a white marble ) = 79  

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Question 30:

A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball?
(a) 25
(b) 35
(c) 110
(d) none of these

Answer:

(b)35

Explanation:
Total possible outcomes = Total number of balls  = ( 4 + 6) = 10  
 Number of black balls = 6
   ∴ (getting a black ball )  =  610= 35  

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Question 31:

A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is not black?
(a) 1815
(b) 215
(c) 1315
(d) 13

Answer:

(c)1315  

Explanation:
Total possible outcomes = Total number of balls

                                      = ( 8 + 2 + 5) = 15
     Total number of non-black balls = (8 + 5) = 13
   ∴ (getting a ball that is not black)  = 1315  

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Question 32:

A bag contains 3 white, 4 red and 5 black balls. One ball is drawn at random. What is the probability that the ball drawn is neither black nor white?
(a) 14
(b) 12
(c) 13
(d) 34

Answer:

(c) 13

Explanation:
Total possible outcomes = Total number of balls

                                      = ( 3 + 4 + 5) = 12 
Total number of balls that are non-black and non-white  = 4
   ∴ (getting a ball that is neither black nor white)  =  412=13  

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Question 33:

A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black king?
(a) 113
(b) 126
(c) 239
(d) none of these

Answer:

(b) 126

Explanation:
Total number of all possible outcomes = 52
Number of  black kings = 2

 ∴ P( getting a black king) =  252 = 126

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Question 34:

From a well-shuffled deck of 52 cards, one card is drawn at random. What is the probability of getting a queen?
(a) 113
(b) 126
(c) 439
(d) none of these

Answer:

(a) 113

Explanation:
Total number of all possible outcomes= 52
Number of queens = 4

 ∴ P( getting a queen) =  452 = 113

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Question 35:

One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a face card?
(a) 126
(b) 326
(c) 313
(d) 413

Answer:

(c) 313

Explanation:
Total number of all possible outcomes= 52
Number of  face cards ( 4 kings + 4 queens  + 4 jacks) = 12

 ∴ P( getting a face card) =  1252 = 313



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Question 36:

Once card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black face card?
(a) 126
(b) 326
(c) 313
(d) 314

Answer:

(b) 326

Explanation:
Total number of all possible outcomes= 52
Number of  black face cards ( 2 kings + 2 queens  + 2 jacks) = 6

 ∴ P( getting a black face card) =  652 = 326

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Question 37:

One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a 6?
(a) 326
(b) 152
(c) 113
(d) none of these

Answer:

(c) 113

Explanation:
Total number of all possible outcomes = 52
Number of 6 in a deck of 52 cards  = 4

 ∴ P( getting a 6) =  452 = 113



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