Rs Aggarwal 2018 for Class 10 Math Chapter 15

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

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 = P (T) = $\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{1}{2}$

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) =

\frac{Number of favourable outcomes}{Total number of possible outcomes} = \frac{2}{4} = \frac{1}{2}

(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) =

\frac{Number of favourable outcomes}{Total number of possible outcomes} = \frac{3}{4}

(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) =

\frac{Number of favourable outcomes}{Total number of possible outcomes} = \frac{3}{4}

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 = P (E) = $\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{3}{6} = \frac{1}{2}$
(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 = P (E) = $\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{4}{6} = \frac{2}{3}$

(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 = P (E) = $\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{4}{6} = \frac{2}{3}$
(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 = P (E) =

\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{2}{6} = \frac{1}{3}

(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 = P (E) =

\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{5}{6}

(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 = P (E) =

\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{1}{6}

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) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{21}{26}$

Thus, the probability of getting a consonant is $\frac{21}{26}$ .

Question 6:

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

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

Answer:

In a single throw of a die, the possible outcomes are:
A,B,C,A,A,B
Total number of possible outcomes = 6
(i) Let E₁ be the event of getting A.
Number of favourable outcomes = 3
∴ P( getting A) = P (E₁) = $\frac{3}{6} = \frac{1}{2}$

(ii) Let E₂ be the event of getting B.
Number of favourable outcomes = 2
∴ P( getting B) = P (E₂) = $\frac{2}{6} = \frac{1}{3}$

Question 7:

It is known that a box of 200 electric bulbs contains 16 defective bulbs. One bulb is taken out at random from the box. What is the probability that the bulb drawn is
(i) defective
(ii) non-defective

Answer:

Total number of possible outcomes = 200

(i) Number of defective bulbs = 16
∴ P(getting a defective bulb) = $\frac{16}{200} = \frac{2}{25}$

(ii) Number of non-defective bulbs = 200 − 16 = 184
∴ P(getting a non-defective bulb) = $\frac{184}{200} = \frac{23}{25}$

Question 8:

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.

Question 9:

There are 35 students in a class of whom 20 are boys and 15 are girls. From these students one is chosen at random. What is the probability that the chosen student is a (i) boy, (ii) girl?

Answer:

Total number of students = 35
Number of boys = 20
Number of girls = 15

(i) Let E₁ be the event that the chosen student is a boy.

∴ P(choosing a boy) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{20}{35} = \frac{4}{7}$

Thus, the probability that the chosen student is a boy is $\frac{4}{7}$ .

(ii) Let E₂ be the event that the chosen student is a girl.

∴ P(choosing a girl) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{15}{35} = \frac{3}{7}$

Thus, the probability that the chosen student is a girl is $\frac{3}{7}$ .

Question 10:

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) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{10}{35} = \frac{2}{7}$

Thus, the probability of getting a prize is $\frac{2}{7}$ .

Question 11:

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
∴ P (Kunal wins a prize) = $\frac{5}{250} = \frac{1}{50}$

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) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{9}{17}$

Thus, the probability that the card drawn bears an odd number is $\frac{9}{17}$ .

(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) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{3}{17}$

Thus, the probability that the card drawn bears a number divisible by 5 is $\frac{3}{17}$ .

Question 13:

A game of chance consists of spinning an arrow, which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. Find the probability that the arrow will point at any factor of 8.

Answer:

Number of all possible outcomes = 8

Let E be the event of getting any factor of 8.

These numbers are 1, 2, 4 and 8.
∴ P(arrow will point at any factor of 8) = P(E) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{4}{8} = \frac{1}{2}$

Thus, the probability that the arrow will point at any factor of 8 is $\frac{1}{2}$ .

Question 14:

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) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{7}{8}$

Thus, the probability of having at least one boy is $\frac{7}{8}$ .

Question 15:

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{2}{15}$

Thus, the probability of getting a black ball is $\frac{2}{15}$ .

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

∴  P(getting a ball which is not green) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{11}{15}$

Thus, the probability of getting a ball which is not green is $\frac{11}{15}$ .

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

∴  P(getting a ball which is red or white) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{9}{15} = \frac{3}{5}$

Thus, the probability of getting a ball which is red or white is $\frac{3}{5}$ .

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

∴  P(getting a ball which is neither red nor green) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{6}{15} = \frac{2}{5}$

Thus, the probability of getting a ball which is neither red nor green is $\frac{2}{5}$ .

Question 16:

A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting

(i) a red king,
(ii) a queen or a jack.

Answer:

Total number of cards = 52

(i) Number of red kings = 2

∴ P(getting a red king) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{2}{52} = \frac{1}{26}$

Thus, the probability of getting a red king is $\frac{1}{26}$ .

(ii) Number of queens or jacks = 4 + 4 = 8

∴ P(getting a queen or a jack) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{8}{52} = \frac{2}{13}$

Thus, the probability of getting a queen or a jack is $\frac{2}{13}$ .

Question 17:

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

Answer:

Favourable outcomes = 52 − 4kings − 4queens = 44

Total outcomes = 52

$P (E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes} = \frac{44}{52} = \frac{11}{13}$

Thus, the probability that the drawn card is neither a king nor a queen is $\frac{11}{13}$ .

Question 18:

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting

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

Answer:

(i) Favourable outcomes = 2red kings + 2red queens + 2red jack = 6

Total outcomes = 52

$P (E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes} = \frac{6}{52} = \frac{3}{26}$

Thus, the probability of getting a red face card is $\frac{3}{26}$ .

(ii) Favourable outcomes = 2 (because there are 4 kings, 2 black and 2 red)

Total outcomes = 52

$P (E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes} = \frac{2}{52} = \frac{1}{26}$

Thus, the probability of getting a black king is $\frac{1}{26}$ .

Question 19:

Two different dice are tossed together. Find the probability that

(i) the number on each die is even,
(ii) the sum of the numbers appearing on the two dice is 5.

Answer:

When two different dice are thrown, then total number of outcomes = 36.

(i) Let E₁ be the event of getting an even number on each die.

These numbers are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4) and (6, 6).

Number of Favourable outcomes = 9

∴ P(getting an even number on both dice) = $P (E_{1}) = \frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$

= $\frac{9}{36} = \frac{1}{4}$

Thus, the probability of getting an even number on both dice is $\frac{1}{4}$ .

(ii) Let E₂ be the event of getting the sum of the numbers appearing on the two dice is 5.

These numbers are (1, 4), (2, 3), (3, 2) and (4, 1)

Number of Favourable outcomes = 4

∴ P(getting the sum of the numbers appearing on the two dice is 5) = $P (E_{2}) = \frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$

= $\frac{4}{36} = \frac{1}{9}$

Thus, the probability of getting the sum of the numbers appearing on the two dice is 5 is $\frac{1}{9}$ .

Question 20:

Two different dice are rolled simultaneously. Find the probability that the sum of the numbers on the two dice is 10.

Answer:

When two different dice are thrown, then total number of outcomes = 36.

Let E₁ be the event of getting the sum of the numbers on the two dice is 10.

These numbers are (4, 6), (5, 5) and (6, 4).

Number of Favourable outcomes = 3

∴ P(getting the sum of the numbers on the two dice is 10) = $P (E_{1}) = \frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$

= $\frac{3}{36} = \frac{1}{12}$

Thus, the probability of getting the sum of the numbers on the two dice is 10 is $\frac{1}{12}$ .

Question 21:

When two dice are tossed together, find the probability that the sum of numbers on their tops is less than 7.

Answer:

When two different dice are thrown, then total number of outcomes = 36.

Let E be the event of getting the sum of the numbers less than 7.

These numbers 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) and (5, 1).

Number of Favourable outcomes = 15

∴ P(getting the sum of the numbers less than 7) = $P (E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes}$

= $\frac{15}{36} = \frac{5}{12}$

Thus, the probability of getting the sum of the numbers less than 7 is $\frac{5}{12}$ .

Question 22:

Two dice are rolled together. Find the probability of getting such numbers on two dice whose product is a perfect square.

Answer:

When two different dice are thrown, then total number of outcomes = 36.

Let E be the event of getting the product of numbers, as a perfect square.

These numbers are (1, 1), (1, 4), (2, 2), (3, 3), (4, 1), (4, 4), (5, 5) and (6, 6).

Number of Favourable outcomes = 8

∴ P(getting the product of numbers, as a perfect square) = $P (E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes}$

= $\frac{8}{36} = \frac{2}{9}$

Thus, the probability of getting the product of numbers, as a perfect square is $\frac{2}{9}$ .

Question 23:

Two dice are rolled together. Find the probability of getting such numbers on the two dice whose product is 12.

Answer:

Number of all possible outcomes = 36

Let E be the event of getting all those numbers whose product is 12.

These numbers are (2, 6), (3, 4), (4, 3) and (6, 2).

∴ P(getting all those numbers whose product is 12) = P(E) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{4}{36} = \frac{1}{9}$

Thus, the probability of getting all those numbers whose product is 12 is $\frac{1}{9}$ .

Question 24:

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) =

\frac{2}{46} = \frac{1}{23}

(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) =

\frac{5}{46}

Question 25:

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?

Answer:

The possible outcomes are 1,2, 3,4, 5................12.
Number of all possible outcomes = 12

(i) Let E₁ be the event that the pointer rests on 6.
Then, number of favourable outcomes = 1
∴ P (arrow pointing at 6) = P( E₁) = $\frac{1}{12}$

(ii) Out of the given numbers, the even numbers are

2, 4, 6, 8,10 and 12

Let E₂ be the event of getting an even number.
Then, number of favourable outcomes = 6
∴ P (arrow pointing at an even number) = P( E₂) =

\frac{6}{12} = \frac{1}{2}

(iii) Out of the given numbers, the prime numbers are 2, 3, 5, 7 and 11.
   Let E₃ be the event of the arrow pointing at a prime number.
  Then, number of favourable outcomes = 5
    ∴ P (arrow pointing at a prime number) = P( E₃)   =

\frac{5}{12}

(iv) Out of the given numbers, the numbers that are multiples of 5 are 5 and 10 only.
Let E₄ be the event of the arrow pointing at a multiple of 5.

Then, number of favourable outcomes = 2
∴ P(arrow pointing at a number that is a multiple of 5) = P( E₄) =

\frac{2}{12} = \frac{1}{6}

Question 26:

12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. One pen is taken out at random from this lot. Find the probability that the pen taken out is good one.

Answer:

Total number of pens = 132 + 12 = 144

Number of good pens = 132

Let E be the event of getting a good pen.

∴ P(getting a good pen) = P(E) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{132}{144} = \frac{11}{12}$

Thus, the probability of getting a good pen is $\frac{11}{12}$ .

Question 27:

A lot consists of 144 ballpoint pens of which 20 are defective and others good. Tanvy will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) she will buy it,
(ii) she will not buy it?

Answer:

Total number of pens = 144
Number of defective pens = 20
Number of good pens = 144 − 20 = 124

(i) Let E₁ be the event of getting a good pen.

∴ P(buying a pen) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{124}{144} = \frac{31}{36}$

Thus, the probability that Tanvy will buy a pen is $\frac{31}{36}$ .

(ii) Let E₂ be the event of getting a defective pen.

∴ P(not buying a pen) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{20}{144} = \frac{5}{36}$

Thus, the probability that Tanvy will not buy a pen is $\frac{5}{36}$ .

Question 28:

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 perfect square number,
(iii) a number divisible by 5.

Answer:

Total number of discs = 90

(i) Let E₁ be the event of having a two-digit number.

Number of discs bearing two-digit number = 90 − 9 = 81

∴ P(getting a two-digit number) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{81}{90} = \frac{9}{10}$

Thus, the probability that the disc bears a two-digit number is $\frac{9}{10}$ .

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

Discs bearing perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64 and 81.

Number of discs bearing a perfect square number = 9

∴ P(getting a perfect square number) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{9}{90} = \frac{1}{10}$

Thus, the probability that the disc bears a perfect square number is $\frac{1}{10}$ .

(iii) Let E₃ be the event of getting a number divisible by 5.

Discs bearing numbers divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90.

Number of discs bearing a number divisible by 5 = 18

∴ P(getting a number divisible by 5) = P(E₃) = $\frac{Number of outcomes favourable to E_{3}}{Number of all possible outcomes}$
= $\frac{18}{90} = \frac{1}{5}$

Thus, the probability that the disc bears a number divisible by 5 is $\frac{1}{5}$ .

Question 29:

(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and not replaced. Now, bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer:

(i) Number of all possible outcomes = 20.
Number of defective bulbs = 4.
Number of non-defective bulbs = 20 − 4 = 16.

Let E₁ be the event of getting a defective bulb.

∴ P(getting a defective bulb) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{4}{20} = \frac{1}{5}$

Thus, the probability that the bulb is defective is $\frac{1}{5}$ .

(ii) After removing 1 non-defective bulb, we have number of remaining bulbs = 19.
Out of these, number of non-defective bulbs = 16 − 1 = 15.

Let E₂ be the event of getting a non-defective bulb.

∴ P(getting a non-defective bulb) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{15}{19}$

Thus, the probability that the bulb is non-defective is $\frac{15}{19}$ .

Question 30:

A bag contains lemon-flavoured candies only. Hema takes out one candy without looking into the bag. What is the probability that she takes out

(i) an orange-flavoured candy?
(ii) a lemon-flavoured candy?

Answer:

Suppose there are x candies in the bag.
Then, number of orange candies in the bag = 0.
And, number of lemon candies in the bag = x.

(i) Let E₁ be the event of getting an orange-flavoured candy.

∴ P(getting an orange-flavoured candy) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{0}{x} = 0$

Thus, the probability that Hema takes out an orange-flavoured candy is 0.

(ii) Let E₂ be the event of getting a lemon-flavoured candy.

∴ P(getting a lemon-flavoured candy) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{x}{x} = 1$

Thus, the probability that Hema takes out a lemon-flavoured candy is 1.

Question 31:

There are 40 students in a class of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. He writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of

(i) a girl?
(ii) a boy?

Answer:

Total number of students = 40.
Number of boys = 15.
Number of girls = 25.

(i) Let E₁ be the event of getting a girl's name on the card.

∴ P(selecting the name of a girl) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{25}{40} = \frac{5}{8}$

Thus, the probability that the name written on the card is the name of a girl is $\frac{5}{8}$ .

(ii) Let E₂ be the event of getting a boy's name on the card.

∴ P(selecting the name of a boy) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{15}{40} = \frac{3}{8}$

Thus, the probability that the name written on the card is the name of a boy is $\frac{3}{8}$ .

Question 32:

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing
(i) an ace
(ii) a 4 of spades
(iii) a 9 of a black suit
(iv) a red king

Answer:

Total number of all possible outcomes= 52
(i) Total number of aces = 4
∴ P( getting an ace) = $\frac{4}{52} = \frac{1}{13}$

(ii) Number of 4 of spades = 1
∴ P(getting a 4 of spade) = $\frac{1}{52}$

(iii) Number of 9 of a black suit = 2
∴ P(getting a 9 of a black suit) = $\frac{2}{52} = \frac{1}{26}$

(iv) Number of red kings = 2
∴ P(getting a red king) = $\frac{2}{52} = \frac{1}{26}$

Question 33:

A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a queen
(ii) a diamond
(iii) a king or an ace
(iv) a red ace

Answer:

Total number of all possible outcomes= 52
(i) Total number of queens = 4
∴ P(getting a queen) = $\frac{4}{52} = \frac{1}{13}$

(ii) Number of diamond suits = 13
∴ P(getting a diamond) = $\frac{13}{52} = \frac{1}{4}$

(iii) Total number of kings = 4
Total number of aces = 4
Let E be the event of getting a king or an ace card.
Then, the favourable outcomes = 4 + 4 = 8
∴ P( getting a king or an ace) = P (E) = $\frac{8}{52} = \frac{2}{13}$

(iv) Number of red aces = 2
∴ P( getting a red ace) = $\frac{2}{52} = \frac{1}{26}$

Question 34:

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red suit
(ii) a face card
(iii) a red face card
(iv) a queen of black suit
(v) a jack of hearts
(vi) a spade

Answer:

Total number of outcomes = 52

(i) Let E₁ be the event of getting a king of red suit.

Number of favourable outcomes = 2

∴ P(getting a king of red suit) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{2}{52} = \frac{1}{26}$

Thus, the probability of getting a king of red suit is $\frac{1}{26}$ .

(ii) Let E₂ be the event of getting a face card.

Number of favourable outcomes = 12

∴ P(getting a face card) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{12}{52} = \frac{3}{13}$

Thus, the probability of getting a face card is $\frac{3}{13}$ .

(iii) Let E₃ be the event of getting a red face card.

Number of favourable outcomes = 6

∴ P(getting a red face card) = P(E₃) = $\frac{Number of outcomes favourable to E_{3}}{Number of all possible outcomes}$
= $\frac{6}{52} = \frac{3}{26}$

Thus, the probability of getting a red face card is $\frac{3}{26}$ .

(iv) Let E₄ be the event of getting a queen of black suit.

Number of favourable outcomes = 2

∴ P(getting a queen of black suit) = P(E₄) = $\frac{Number of outcomes favourable to E_{4}}{Number of all possible outcomes}$
= $\frac{2}{52} = \frac{1}{26}$

Thus, the probability of getting a queen of black suit is $\frac{1}{26}$ .

(v) Let E₅ be the event of getting a jack of hearts.

Number of favourable outcomes = 1

∴ P(getting a jack of hearts) = P(E₅) = $\frac{Number of outcomes favourable to E_{5}}{Number of all possible outcomes}$
= $\frac{1}{52}$

Thus, the probability of getting a jack of hearts is $\frac{1}{52}$ .

(vi) Let E₆ be the event of getting a spade.

Number of favourable outcomes = 13

∴ P(getting a spade) = P(E₆) = $\frac{Number of outcomes favourable to E_{6}}{Number of all possible outcomes}$
= $\frac{13}{52} = \frac{1}{4}$

Thus, the probability of getting a spade is $\frac{1}{4}$ .

Question 35:

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) =

\frac{16}{52} = \frac{4}{13}

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

\frac{2}{52} = \frac{1}{26}

(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) =

\frac{8}{52} = \frac{2}{13}

(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 ) =

\frac{2}{13}

Now, P (E) + P (not E) = 1
∴ P(getting neither a king nor a queen ) =

1 - (\frac{2}{13}) = \frac{11}{13}

Question 1:

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) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{16}{25}$

Thus, the probability that the number on the drawn card is divisible by 2 or 3 is $\frac{16}{25}$ .

(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) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{9}{25}$

Thus, the probability that the number on the drawn card is a prime number is $\frac{9}{25}$ .

Question 2:

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 T_n= 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) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{11}{18}$

Thus, the probability that the number on the card is a prime number is $\frac{11}{18}$ .

Question 3:

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) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{20}{30} = \frac{2}{3}$

Thus, the probability that the number on the card is not divisible by 3 is $\frac{2}{3}$ .

(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) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{6}{30} = \frac{1}{5}$

Thus, the probability that the number on the card is a prime number greater than 7 is $\frac{1}{5}$ .

(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) = $\frac{Number of outcomes favourable to E_{3}}{Number of all possible outcomes}$
= $\frac{25}{30} = \frac{5}{6}$

Thus, the probability that the number on the card is not a perfect square number is $\frac{5}{6}$ .

Question 4:

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 T_n = 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) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{5}{18}$

Thus, the probability of getting a card bearing a prime number less than 15 is $\frac{5}{18}$ .

(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) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{1}{18}$

Thus, the probability of getting a card bearing a number divisible by 3 and 5 is $\frac{1}{18}$ .

Question 5:

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 T_n = 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) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{4}{65}$

Thus, the probability that the card bears a one-digit number is $\frac{4}{65}$ .

(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 T_n = 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) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{13}{65} = \frac{1}{5}$

Thus, the probability that the card bears a number divisible by 5 is $\frac{1}{5}$ .

(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 T_n = 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) = $\frac{Number of outcomes favourable to E_{3}}{Number of all possible outcomes}$
= $\frac{12}{65}$

Thus, the probability that the card bears an odd number less than 30 is $\frac{12}{65}$ .

(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) = $\frac{Number of outcomes favourable to E_{4}}{Number of all possible outcomes}$
= $\frac{15}{65} = \frac{3}{13}$

Thus, the probability that the card bears a composite number between 50 and 70 is $\frac{3}{13}$ .

Question 6:

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 T_n = 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 T_n = 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) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{9}{51} = \frac{3}{17}$

Thus, the probability that the number on the drawn card is less than 19 is $\frac{3}{17}$ .

(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) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{7}{51}$

Thus, the probability that the number on the drawn card is a prime number less than 20 is $\frac{7}{51}$ .

Question 7:

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    $[T_{n} = 100 \Rightarrow 2 + (n - 1) \times 2 = 100, \Rightarrow n = 50]$
∴ P (getting an even number) = $\frac{50}{100} = \frac{1}{2}$

(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) = $\frac{14}{100} = \frac{7}{50}$

(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) =

\frac{9}{100}

(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) =

\frac{12}{100} = \frac{3}{25}

Question 8:

A box contains 80 discs, which are numbered from 1 to 80. If one disc is drawn at random from the box, find the probability that it bears a perfect square number.

Answer:

Total number of outcomes = 80.

Let E₁ be the event of getting a perfect square number.

Out of these numbers, perfect square numbers are 1, 4, 9, 16, 25, 36, 49 and 64.

Thus, number of favourable outcomes = 8.

∴ P(getting a perfect square number) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{8}{80} = \frac{1}{10}$

Thus, the probability that the disc bears a perfect square number is $\frac{1}{10}$ .

Question 9:

A piggy bank contains hundred 50 p coins, seventy ₹1 coin, fifty ₹2 coins and thirty ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin

(i) will be a ₹1 coin?
(ii) will not be a ₹5 coin?
(iii) will be a 50 p or a ₹2 coin?

Answer:

Number of 50 p coins = 100.
Number of ₹1 coins = 70.
Number of ₹2 coins = 50.
Number of ₹5 coins = 30.

Thus, total number of outcomes = 250.

(i) Let E₁ be the event of getting a ₹1 coin.

Number of favourable outcomes = 70.

∴ P(getting a ₹1 coin) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{70}{250} = \frac{7}{25}$

Thus, the probability that the coin will be a ₹1 coin is $\frac{7}{25}$ .

(ii) Let E₂ be the event of not getting a ₹5 coin.

Number of favourable outcomes = 250 − 30 = 220

∴ P(not getting a ₹5 coin) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{220}{250} = \frac{22}{25}$

Thus, the probability that the coin will not be a ₹5 coin is $\frac{22}{25}$ .

(iii) Let E₃ be the event of getting a 50 p or a ₹2 coin.

Number of favourable outcomes = 100 + 50 = 150

∴ P(getting a 50 p or a ₹2 coin) = P(E₃) = $\frac{Number of outcomes favourable to E_{3}}{Number of all possible outcomes}$
= $\frac{150}{250} = \frac{3}{5}$

Thus, the probability that the coin will be a 50 p or a ₹2 coin is $\frac{3}{5}$ .

Question 10:

The probability of selecting a red ball at random from the jar that contains only red, blue and orange balls is $\frac{1}{4}$ . The probability of selecting a blue ball at random from the same jar is $\frac{1}{3}$ . 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) = $\frac{1}{4}$ and P(getting a blue ball) = $\frac{1}{3}$

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.

$∴ \frac{1}{4} + \frac{1}{3} + x = 1 \Rightarrow x = 1 - \frac{1}{4} - \frac{1}{3} \Rightarrow x = \frac{12 - 3 - 4}{12} \Rightarrow x = \frac{5}{12}$ \
∴ P(getting an orange ball) = $\frac{5}{12}$

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

∴ P(getting an orange ball) = $\frac{10}{n}$
$\Rightarrow \frac{10}{n} = \frac{5}{12} \Rightarrow n = 24$

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

Question 11:

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 $\frac{9}{8}$ 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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{18 - x}{18}$

Thus, the probability of drawing a ball which is not red is $\frac{18 - x}{18}$ .

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

P(getting a red ball now) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{x + 2}{20}$

and P(getting a red ball in first case) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{x}{18}$

Since, it is given that probability of drawing a red ball now will be $\frac{9}{8}$ times the probability of drawing a red ball in the first case.

Thus, $\frac{x + 2}{20} = \frac{9}{8} \times \frac{x}{18}$
$\Rightarrow 144 (x + 2) = 180 x \Rightarrow 144 x + 288 = 180 x \Rightarrow 36 x = 288 \Rightarrow x = \frac{288}{36} = 8$

Thus, the value of x is 8.

Question 12:

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 $\frac{2}{3}$ . 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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{24 - x}{24}$

But, P(getting a green marble) = $\frac{2}{3}$ (given)

$∴ \frac{24 - x}{24} = \frac{2}{3} \Rightarrow 3 (24 - x) = 48 \Rightarrow 72 - 3 x = 48 \Rightarrow 3 x = 72 - 48 \Rightarrow 3 x = 24 \Rightarrow x = 8$

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

Question 13:

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 $\frac{1}{3}$ and the probability of selecting a green marble at random is $\frac{4}{9}$ . How many white marbles does the jar contain?

Answer:

Total number of marbles = 54.

It is given that, P(getting a blue marble) = $\frac{1}{3}$ and P(getting a green marble) = $\frac{4}{9}$

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.

$∴ \frac{1}{3} + \frac{4}{9} + x = 1 \Rightarrow \frac{3 + 4}{9} + x = 1 \Rightarrow x = 1 - \frac{7}{9} \Rightarrow x = \frac{2}{9}$
∴ P(getting a white marble) = $\frac{2}{9}$ ...(1)

Let the number of white marbles be n.

Then, P(getting a white marble) = $\frac{n}{54}$ ... (2)

From (1) and (2),
$\frac{n}{54} = \frac{2}{9} \Rightarrow n = \frac{2 \times 54}{9} \Rightarrow n = 12$

Thus, there are 12 white marbles in the jar.

Question 14:

A carton consists of 100 shirts of which 88 are good and 8 have minor defects. Rohit, a trader, will only accept the shirts which are good. But Kamal, an another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that it is acceptable to

(i) Rohit,
(ii) Kamal?

Answer:

Total number of shirts = 100.
Number of good shirts = 88.
Number of shirts with minor defects = 8.
Number of shirts with major defects = 100 − 88 − 8 = 4.

(i) P(the drawn shirt is acceptable to Rohit) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{88}{100} = \frac{22}{25}$

Thus, the probability that the drawn shirt is acceptable to Rohit is $\frac{22}{25}$ .

(ii) P(the drawn shirt is acceptable to Kamal) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{88 + 8}{100} = \frac{96}{100} = \frac{24}{25}$

Thus, the probability that the drawn shirt is acceptable to Kamal is $\frac{24}{25}$ .

Question 15:

A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is

(i) extremely patient,
(ii) extremely kind or honest.

Which of the above values you prefer more?

Answer:

Total number of persons = 12.
Number of persons who are extremely patient = 3.
Number of persons who are extremely honest = 6.
Number of persons who are extremely kind = 12 − 3 − 6 = 3.

(i) P(selecting a person who is extremely patient) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{3}{12} = \frac{1}{4}$

Thus, the probability of selecting a person who is extremely patient is $\frac{1}{4}$ .

(ii) P(selecting a person who is extremely kind or honest) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{6 + 3}{12} = \frac{9}{12} = \frac{3}{4}$

Thus, the probability of selecting a person who is extremely kind or honest is $\frac{3}{4}$ .

From the three given values, we prefer honesty more.

Question 16:

A die is rolled twice. Find the probability that

(i) 5 will not come up either time,
(ii) 5 will come up exactly one time,
(iii) 5 will come up both the times.

Answer:

Total number of outcomes = 36.

(i) Cases where 5 comes up on at least one time are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 5).

Number of such cases = 11.

Number of cases where 5 will not come up either time = 36 − 11 = 25.

∴ P(5 will not come up either time) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{25}{36}$

Thus, the probability that 5 will not come up either time is $\frac{25}{36}$ .

(ii) Cases where 5 comes up on exactly one time are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) and (6, 5).

Number of such cases = 10.

∴ P(5 will come up exactly one time) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{10}{36} = \frac{5}{18}$

Thus, the probability that 5 will come up exactly one time is $\frac{5}{18}$ .

(iii) Cases where 5 comes up on exactly two times is (5, 5).

Number of such cases = 1.

∴ P(5 will come up both the times) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{36}$

Thus, the probability that 5 will come up both the times is $\frac{1}{36}$ .

Question 17:

Two dice are rolled once. Find the probability of getting such numbers on two dice whose product is a perfect square.

Answer:

Number of possible outcomes = 36

Let E be the event of getting two numbers whose product is a perfect square.
Then, the favourable outcomes are (1, 1), (1, 4), (4, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

Number of favourable outcomes = 8.

∴ P(getting numbers whose product is a perfect square) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{8}{36} = \frac{2}{9}$

Thus, the probability of getting such numbers on two dice whose product is a perfect square is $\frac{2}{9}$ .

Question 18:

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) = $\frac{6}{11}$

(ii) Number of consonants in the given word ( S, S, C, T, N) = 5
∴ P (getting a consonant) = $\frac{5}{11}$

(iii) Number of S in the given word = 2
∴ P (getting an S) = $\frac{2}{11}$

Question 19:

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{5}$

Thus, the probability that the drawn card is the queen is $\frac{1}{5}$ .

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

(i) Number of aces = 1.

∴ P(getting an ace) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{4}$

Thus, the probability that the drawn card is an ace is $\frac{1}{4}$ .

(ii) Number of queens = 0.

∴ P(getting a queen now) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{0}{4} = 0$

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

Question 20:

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) = $\frac{24}{52} = \frac{6}{13}$

Question 21:

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) = $\frac{1}{7}$

Thus, the probability that an ordinary year has 53 Mondays is $\frac{1}{7}$ .

Question 22:

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{20}{46} = \frac{10}{23}$

Thus, the probability that the drawn card is a red card is $\frac{10}{23}$ .

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

∴ P(getting a face card) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{6}{46} = \frac{3}{23}$

Thus, the probability that the drawn card is a face card is $\frac{3}{23}$ .

(iii) Number of card of clubs = 12.

∴ P(getting a card of clubs) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{12}{46} = \frac{6}{23}$

Thus, the probability that the drawn card is a card of clubs is $\frac{6}{23}$ .

Question 23:

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{2}{40} = \frac{1}{20}$

Thus, the probability that the drawn card is a black face card is $\frac{1}{20}$ .

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

∴ P(getting a red card) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{20}{40} = \frac{1}{2}$

Thus, the probability that the drawn card is a red card is $\frac{1}{2}$ .

Question 24:

A game consists of tossing a one-rupee coin three times, and noting its outcome each time. Find the probability of getting

(i) three heads,
(ii) at least two tails.

Answer:

When a coin is tossed three times, all possible outcomes are
HHH, HHT, HTH, THH, HTT, THT, TTH and TTT.

Number of total outcomes = 8.

(i) Outcome with three heads is HHH.

Number of outcomes with three heads = 1.

∴ P(getting three heads) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{8}$

Thus, the probability of getting three heads is $\frac{1}{8}$ .

(ii) Outcomes with atleast two tails are TTH, THT, HTT and TTT.

Number of outcomes with atleast two tails = 4.

∴ P(getting at least two tails) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{4}{8} = \frac{1}{2}$

Thus, the probability of getting at least two tails is $\frac{1}{2}$ .

Question 25:

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{2}{7}$

Thus, the probability that a leap year selected at random will contain 53 Sundays is $\frac{2}{7}$ .

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).

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) $(1 - \frac{1}{p})$

Answer:

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

Hence, the correct answer is option (b).

Question 3:

What is the probability of an impossible event?
(a) $\frac{1}{2}$
(b) 0
(c) 1
(d) none of these

Answer:

(b) 0

The probability of an impossible event is 0.

Question 4:

What is the probability of a sure event?
(a) 0
(b) $\frac{1}{2}$
(c) 1
(d) none of these

Answer:

(c) 1

The probability of a sure event is 1.

Question 5:

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

(a) 1.5
(b) $\frac{3}{5}$
(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).

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) $\frac{2}{3}$
(b) $\frac{1}{6}$
(c) $\frac{1}{3}$
(d) $\frac{11}{30}$

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{10}{30} = \frac{1}{3}$

Thus, the probability that the selected number is a prime number is $\frac{1}{3}$ .

Hence, the correct answer is option (c).

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) $\frac{4}{15}$
(b) $\frac{2}{15}$
(c) $\frac{1}{5}$
(d) $\frac{1}{3}$

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{3}{15} = \frac{1}{5}$

Thus, the probability that a number selected is a multiple of 4 is $\frac{1}{5}$ .

Hence, the correct answer is option (c).

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) $\frac{1}{45}$
(b) $\frac{2}{15}$
(c) $\frac{4}{45}$
(d) $\frac{1}{9}$

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{5}{45} = \frac{1}{9}$

Thus, the probability that the drawn card has a number which is a perfect square is $\frac{1}{9}$ .

Hence, the correct answer is option (d).

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) $\frac{7}{90}$
(b) $\frac{1}{9}$
(c) $\frac{4}{15}$
(d) $\frac{8}{89}$

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{8}{90} = \frac{4}{45}$

Thus, the probability that the disc bears prime number less than 23 is $\frac{4}{45}$ .

Disclaimer: There is a misprinting in the question. If they ask for the probability of all prime numbers less than 90, then we get $\frac{4}{15}$ .

Hence, the correct answer is option (c).

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) $\frac{1}{2}$
(b) $\frac{2}{5}$
(c) $\frac{3}{10}$
(d) $\frac{5}{9}$

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{5}{10} = \frac{1}{2}$

Thus, the probability of getting a card with a prime number is $\frac{1}{2}$ .

Hence, the correct answer is option (a).

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) $\frac{1}{7}$
(b) $\frac{1}{8}$
(c) $\frac{1}{5}$
(d) $\frac{7}{40}$

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{5}{40} = \frac{1}{8}$

Thus, the probability that the selected ticket has a number, which is a multiple of 7, is $\frac{1}{8}$ .

Hence, the correct answer is option (b).

Question 12:

Which of the following cannot be the probability of an event?
(a) $\frac{1}{3}$
(b) 0.3
(c) 33%
(d) $\frac{7}{6}$

Answer:

(d) $\frac{7}{6}$

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

Question 13:

If the probability of winning a game is 0.4, the probability of losing it is
(a) 0.96
(b) $\frac{1}{0.4}$
(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

Question 14:

If an event cannot occur then its probability is
(a) 1
(b) $\frac{1}{2}$
(c) $\frac{3}{4}$
(d) 0

Answer:

(d) 0

If an event cannot occur, its probability is 0.

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) $\frac{1}{4}$
(b) $\frac{1}{5}$
(c) $\frac{2}{5}$
(d) $\frac{3}{10}$

Answer:

(b) $\frac{1}{5}$

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) = $\frac{4}{20} = \frac{1}{5}$

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) $\frac{2}{5}$
(b) $\frac{11}{25}$
(c) $\frac{12}{25}$
(d) $\frac{13}{25}$

Answer:

(c) $\frac{12}{25}$

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
∴ P (getting a multiple of 3 or 5 ) = P (E) = $\frac{12}{25}$

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) $\frac{3}{5}$
(b) $\frac{1}{3}$
(c) $\frac{1}{2}$
(d) $\frac{2}{5}$

Answer:

(d) $\frac{2}{5}$

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) =

\frac{4}{10} = \frac{2}{5}

Question 18:

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

(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{1}{6}$
(d) $\frac{5}{6}$

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{3}{6} = \frac{1}{2}$

Thus, the probability of getting an even number is $\frac{1}{2}$ .

Hence, the correct answer is option (a).

Question 19:

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

(a) $\frac{2}{5}$
(b) $\frac{5}{6}$
(c) $\frac{1}{3}$
(d) $\frac{2}{3}$

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{4}{6} = \frac{2}{3}$

Thus, the probability of getting a number greater than 2 is $\frac{2}{3}$ .

Hence, the correct answer is option (d).

Question 20:

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

(a) $\frac{1}{3}$
(b) $\frac{1}{6}$
(c) $\frac{1}{2}$
(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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{6}$

Thus, the probability of getting an odd number greater than 3 is $\frac{1}{6}$ .

Hence, the correct answer is option (b).

Question 21:

A die is thrown once. The probability of getting a prime number is
(a) $\frac{2}{3}$
(b) $\frac{1}{3}$
(c) $\frac{1}{2}$
(d) $\frac{1}{6}$

Answer:

(c) $\frac{1}{2}$

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 = P (E) = $\frac{Number of favorable outcomes}{Total number of possible outcomes} = \frac{3}{6} = \frac{1}{2}$

Question 22:

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

(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{1}{6}$
(d) $\frac{1}{12}$

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{6}{36} = \frac{1}{6}$

Thus, the probability of getting the same number on both dice is $\frac{1}{6}$ .

Hence, the correct answer is option (c).

Question 23:

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

(a) 1
(b) $\frac{3}{4}$
(c) $\frac{1}{2}$
(d) $\frac{1}{4}$

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) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{4}$

Thus, the probability of getting 2 heads is $\frac{1}{4}$ .

Hence, the correct answer is option (d).

Question 24:

Two dice are thrown simultaneously. What is the probability of getting a doublet?
(a) $\frac{1}{36}$
(b) $\frac{5}{12}$
(c) $\frac{1}{6}$
(d) $\frac{2}{3}$

Answer:

(c) $\frac{1}{6}$

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) = $\frac{6}{36} = \frac{1}{6}$

Question 25:

Two coins are tossed simultaneously. What is the probability of getting at most one head?
(a) $\frac{1}{4}$
(b) $\frac{1}{2}$
(c) $\frac{2}{3}$
(d) $\frac{3}{4}$

Answer:

(d) $\frac{3}{4}$

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) =

\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{3}{4}

Question 26:

Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
(i) $\frac{1}{2}$
(ii) $\frac{1}{4}$
(iii) $\frac{3}{8}$
(iv) $\frac{3}{4}$

Answer:

(c) $\frac{3}{8}$

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) =

\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{3}{8}

Question 27:

In a lottery, there are 8 prizes and 16 blanks. What is the probability of getting a prize?
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{2}{3}$
(d) none of these

Answer:

(b) $\frac{1}{3}$

Explanation:
Number of prizes = 8
Number of blanks = 16
Total number of tickets = 8 +16= 24
∴ P(getting a prize ) = $\frac{8}{24} = \frac{1}{3}$

Question 28:

In a lottery, there are 6 prizes and 24 blanks. What is the probability of not getting a prize?
(a) $\frac{3}{4}$
(b) $\frac{3}{5}$
(c) $\frac{4}{5}$
(d) none of these

Answer:

(c) $\frac{4}{5}$

Explanation:
Number of prizes = 6
Number of blanks = 24
Total number of tickets = 6 + 24= 30
∴ P(not getting a prize ) = $\frac{24}{30} = \frac{4}{5}$

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) $\frac{1}{3}$
(b) $\frac{4}{9}$
(c) $\frac{7}{9}$
(d) $\frac{2}{9}$

Answer:

(c) $\frac{7}{9}$

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 ) =

\frac{7}{9}

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) $\frac{2}{5}$
(b) $\frac{3}{5}$
(c) $\frac{1}{10}$
(d) none of these

Answer:

(b) $\frac{3}{5}$

Explanation:
Total possible outcomes = Total number of balls = ( 4 + 6) = 10
Number of black balls = 6
∴ P (getting a black ball ) = $\frac{6}{10} = \frac{3}{5}$

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) $\frac{18}{15}$
(b) $\frac{2}{15}$
(c) $\frac{13}{15}$
(d) $\frac{1}{3}$

Answer:

(c) $\frac{13}{15}$

Explanation:
Total possible outcomes = Total number of balls

= ( 8 + 2 + 5) = 15

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

\frac{13}{15}

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) $\frac{1}{4}$
(b) $\frac{1}{2}$
(c) $\frac{1}{3}$
(d) $\frac{3}{4}$

Answer:

(c) $\frac{1}{3}$

Explanation:
Total possible outcomes = Total number of balls

= ( 3 + 4 + 5) = 12

Total number of balls that are non-black and non-white = 4
∴ P (getting a ball that is neither black nor white) =

\frac{4}{12} = \frac{1}{3}

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) $\frac{1}{13}$
(b) $\frac{1}{26}$
(c) $\frac{2}{39}$
(d) none of these

Answer:

(b) $\frac{1}{26}$

Explanation:
Total number of all possible outcomes = 52
Number of black kings = 2
∴ P( getting a black king) = $\frac{2}{52} = \frac{1}{26}$

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) $\frac{1}{13}$
(b) $\frac{1}{26}$
(c) $\frac{4}{39}$
(d) none of these

Answer:

(a) $\frac{1}{13}$

Explanation:
Total number of all possible outcomes= 52
Number of queens = 4
∴ P( getting a queen) = $\frac{4}{52} = \frac{1}{13}$

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) $\frac{1}{26}$
(b) $\frac{3}{26}$
(c) $\frac{3}{13}$
(d) $\frac{4}{13}$

Answer:

(c) $\frac{3}{13}$

Explanation:
Total number of all possible outcomes= 52
Number of face cards ( 4 kings + 4 queens + 4 jacks) = 12
∴ P( getting a face card) = $\frac{12}{52} = \frac{3}{13}$

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) $\frac{1}{26}$
(b) $\frac{3}{26}$
(c) $\frac{3}{13}$
(d) $\frac{3}{14}$

Answer:

(b) $\frac{3}{26}$

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) = $\frac{6}{52} = \frac{3}{26}$

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) $\frac{3}{26}$
(b) $\frac{1}{52}$
(c) $\frac{1}{13}$
(d) none of these

Answer:

(c) $\frac{1}{13}$

Explanation:
Total number of all possible outcomes = 52
Number of 6 in a deck of 52 cards = 4
∴ P( getting a 6) = $\frac{4}{52} = \frac{1}{13}$

Rs Aggarwal 2018 for Class 10 Math Chapter 15 - Probability

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