Rs Aggarwal (2015) Solutions for Class 10 Math Chapter 15 Probability are provided here with simple step-by-step explanations. These solutions for Probability are extremely popular among Class 10 students for Math Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal (2015) Book of Class 10 Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal (2015) Solutions. All Rs Aggarwal (2015) Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Page No 609:

#### 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 609:

#### 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*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{1}{2}$

#### Page No 609:

#### Question 3:

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*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{6}=\frac{1}{2}$

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

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

(iv)

Let E be the event of getting a number between 3 and 6.

∴ Probability of getting a number between 3 and 6 =

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{2}{6}=\frac{1}{3}$

(v)

Let E be the event of getting a number other than 3.

∴ Probability of getting a number other than 3 =

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{5}{6}$

(vi)

Let E be the event of getting the number 5.

∴ Probability of getting the number 5 =

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{1}{6}$

#### Page No 609:

#### Question 4:

Two coins are tossed simultaneously. Find the probability of getting

(i) exactly one head

(ii) at most one head

(iii) no tail

#### 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.

â€‹Number of favourable outcomes = 2

*P*(getting exactly 1 head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}u\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{2}{4}=\frac{1}{2}$

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

â€‹Number of favourable outcomes = 3

*P*(getting at most 1 head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{4}$

(iii) Let E be the event of getting no tail.

â€‹Number of favourable outcomes = 1

#### Page No 609:

#### Question 5:

Three coins are tossed simultaneously. What is the probability of getting at least two heads?

#### Answer:

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 at least two heads.

â€‹Number of favourable outcomes = 4

*P*(getting at least 2 heads) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{4}{8}=\frac{1}{2}$

#### Page No 609:

#### Question 6:

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) = $\frac{184}{200}=\frac{23}{25}$

#### Page No 609:

#### Question 7:

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) = $\frac{5}{250}=\frac{1}{50}$

#### Page No 610:

#### Question 8:

A bag contains 5 white, 6 red and 4 green balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) green

(ii) white

(iii) non-red?

#### Answer:

Total number of possible outcomes = Total number of balls

*P*(getting a green ball ) = $\frac{4}{15}$

(ii) Number of white balls = 5

∴

*P*(getting a white ball ) = $\frac{5}{15}=\frac{1}{3}$

(iii) Total number of non-red balls = 9

∴

*P*(getting a non-red ball ) = $\frac{9}{15}=\frac{3}{5}$

#### Page No 610:

#### Question 9:

A bag contains 3 red, 5 black and 7 white balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) white

(ii) red

(iii) not red

(iv) red or white?

#### Answer:

Total possible outcomes = Total number of balls

*P*(getting a white ball) = $\frac{7}{15}$

(ii) Number of red balls = 3

∴

*P*(getting a red ball) = $\frac{3}{15}=\frac{1}{5}$

(iii) Total number of non-red balls = (5 +7) = 12

∴

*P*(getting a non-red ball) = $\frac{12}{15}=\frac{4}{5}$

(iv) â€‹Total number of red or white balls = (3 +7) = 10

∴

*P*(getting a red or white ball) = $\frac{10}{15}=\frac{2}{3}$

#### Page No 610:

#### Question 10:

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 *E*_{1} be the event of getting A.

Number of favourable outcomes = 3

∴ *P*( getting A) = *P *(*E*_{1}) = $\frac{3}{6}=\frac{1}{2}$

(ii) Let *E*_{2} be the event of getting B.

Number of favourable outcomes = 2

∴ *P*( getting B) = *P *(*E*_{2}) = $\frac{2}{6}=\frac{1}{3}$

#### Page No 610:

#### Question 11:

A bag contains 7 red, 5 white and 3 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is

(i) red or white

(ii) not black

(iii) neither white nor black

#### Answer:

Total possible outcomes = Total number of balls

*P*(getting a red or white ball) = $\frac{12}{15}=\frac{4}{5}$

(ii) Total number of non-black balls = (7 + 5) = 12

∴

*P*(getting a not black ball) = $\frac{12}{15}=\frac{4}{5}$

(iii) Number of balls that are neither white nor black = 7

∴

*P*(getting neither a white nor black ball ) = $\frac{7}{15}$

#### Page No 610:

#### Question 12:

A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) white or blue

(ii) red or black

(iii) not white

(iv) neither white nor black

#### Answer:

Total possible outcomes = Total number of balls

*P*(getting a white or blue ball ) = $\frac{7}{18}$

(ii) Total number of red and black balls = (7 + 4) = 11

∴

*P*(getting a red or black ball ) = $\frac{11}{18}$

(iii)Total number of non-white balls = (7 + 4 + 2) = 13

∴

*P*(getting a ball that is not white) = $\frac{13}{18}$

(iv) Total number of balls that are non-white and non-black = (7 + 2) = 9

∴

*P*(getting a ball that is neither white nor black) = $\frac{9}{18}=\frac{1}{2}$

#### Page No 610:

#### Question 13:

A bag contains 5 red, 4 blue and 3 green balls. A ball is taken out of the bag at random. Find the probability that the selected ball is (i) of red colour (ii) not of green colour.

#### Answer:

Total possible outcomes = Total number of balls

*P*(getting a red ball ) = $\frac{5}{12}$

(ii) Total number of balls which are not green = (5 + 4) = 9

∴

*P*(getting a ball which is not green) = $\frac{9}{12}=\frac{3}{4}$

#### Page No 610:

#### Question 14:

A bag contains 4 white and some red balls. If the probability of drawing a red ball is double that of drawing a white ball, find the number of red balls in the bag.

#### Answer:

Let the number of red balls in the bag be *x*.

Then, total number of balls = ( 4 + *x*)

Given: *P*(red ball) = 2 â¨¯ *P*( white ball)

∴ $\frac{x}{4+x}=2\times \left(\frac{4}{4+x}\right)\Rightarrow x=8$

Hence, the number of red balls in the bag is 8.

#### Page No 610:

#### Question 15:

Two dice are thrown simultaneously. What is the probability that

(i) 5 will not come up on either of them?

(ii) 5 will come up on at least one of them?

(iii) 5 will come up at both the dice?

#### Answer:

When two dice are thrown simultaneously, the 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

(i) Let *E*_{1} be the event in which 5 will not come up on either of them.

Then, the favourable outcomes are:

(1,1), (1,2), (1,3), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,6), (3,1), (3,2), (3,3), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,6), (6,1), (6,2), (6,3), (6,4) and (6,6).

Number of favourable outcomes = 25

∴ *P* (5 will not come up on either of them) = *P* (E_{1}) = $\frac{25}{36}$

(ii) Let *E*_{2} be the event in which 5 will come up on at least one.

Then the favourable outcomes are:

(1,5), (2,5), (3,5), (4,5), (5,5), â€‹(6,5), â€‹(5,1), â€‹(5,2), â€‹â€‹(5,3), â€‹â€‹(5,4) and â€‹â€‹(5,6)

Number of favourable outcomes = 11

∴ *P* ( 5 will come up on at least one) = *P* (E_{2}) = $\frac{11}{36}$

(iii) Let *E*_{3} be the event in which 5 will come up on both the dice.

Then the favourable outcome is (5,5).

Number of favourable outcomes = 1

∴ *P* (5 will come up on both the diceâ€‹) = *P* (E_{2}) = $\frac{1}{36}$

#### Page No 610:

#### Question 16:

Find the probability that a number selected at random from the numbers 1 to 25 is not a prime number, when each of the given numbers is equally likely to be selected.

#### Answer:

The possible outcomes are 1,2,3,4,................25.

Number of all possible outcomes = 25

Total prime numbers from 1 to 25 = 9

Thus, if the selected number is not prime, then number of favourable outcomes = 25- 9 = 16

∴

*P*(getting a non-prime number) = $\frac{16}{25}$

#### Page No 610:

#### Question 17:

There are 30 cards numbered from 1 to 30. One card is drawn at random. Find the probability that the number on the selected card is not divisible by 3.

#### Answer:

The possible outcomes are 1,2,3,4,................30.

Number of all possible outcomes = 30

Total numbers divisible by 3 = 10

Let E be the event of getting a number that is not divisible by 3.

*P*(getting a number that is not divisible by 3) =

*P*(E) = $\frac{20}{30}=\frac{2}{3}$

#### Page No 610:

#### Question 18:

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

(i) even

(ii) prime

(iii) multiple of 6

#### Answer:

The possible outcomes are 1,2, 3,4, 5................25.

Number of all possible outcomes = 25

(i) Out of the given numbers, the even numbers are 2, 4, 6, 8,10,12,14,16, 18, 20,22 and 24. Let E_{1} be the event of getting an even number.

Then, number of favourable outcomes = 12â€‹

∴* P* (getting an even number) = $\frac{12}{25}$

(ii) Out of the given numbers, the prime numbers are 2, 3, 5, 7,11,13, 17, 19 and 23. Let E_{2} be the event of getting a prime number.

Then,â€‹ number of favourable outcomes =â€‹ 9

∴* P* (getting a prime number) = $\frac{9}{25}$

(iii) Out of the given numbers, the numbers that are multiples of 6 are 6,12,18 and 24.

Let E_{3} be the event of getting a multiple of 6.

∴

*P*(getting a multiple of 6) = $\frac{4}{25}$

#### Page No 611:

#### Question 19:

A box contains 19 balls bearing numbers 1, 2, 3, ..., 19 respectively. A ball is drawn at random from the box. Find the probability that the number on the ball is

(i) a prime number

(ii) divisible by 3 or 5

(iii) neither divisible by 5 nor by 10

(iv) an even number

#### Answer:

The possible outcomes are 1,2, 3,4, 5................19.

Number of all possible outcomes = 19

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

Let E_{1} be the event of getting a prime number.

Then, number of favourable outcomes = 8â€‹

∴* P* (getting a prime number) = $\frac{8}{19}$

(ii) Out of the given numbers, the numbers divisible by 3 are 3, 6, 9, 12, 15 and 18.

_{2}be the event of getting a number divisible by 3 or 5.

∴

*P*(getting a number divisible by 3 or 5 ) = $\frac{8}{19}$

(iii) Out of the given numbers, the numbers divisible by 5 or by 10 are 5, 10, 15.

Let E

_{3}be the event of getting a number which is neither divisible by 5 nor by 10.

Then, number of favourable outcomes = 19 −3 = 16â€‹

∴

*P*(getting a number which is neither divisible by 5 nor by 10) = $\frac{16}{19}$

(iv) Out of the given numbers , the even numbers are 2, 4, 6, 8, 10, 12, 14, 16 and 18.

_{4}be the event of getting an even number.

Then,â€‹ number of favourable outcomes =â€‹ 9

∴

*P*(getting an even number) = $\frac{9}{19}$

#### Page No 611:

#### Question 20:

A box contains 20 balls bearing numbers 1, 2, 3,..., 20 respectively. A ball is drawn at random from the box. What is the probability that the number on the ball is

(i) an odd number

(ii) divisible by 2 or 3

(iii) a prime number

(iv) not divisible by 10

#### Answer:

The possible outcomes are 1, 2, 3, 4, 5................20.

Number of all possible outcomes = 20

(i) Out of the given numbers, the odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19. Let E_{1} be the event of getting an odd number.

Then, number of favourable outcomes = 10â€‹

∴* P* (getting an odd number) = $\frac{10}{20}=\frac{1}{2}$

(ii) Out of the given numbers, numbers divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20.

_{2}be the event of getting a number divisible by 2 or 3 .

∴

*P*(getting a number divisible by 2 or 3 ) = $\frac{13}{20}$

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

_{3}be the event of getting a prime number.

Then, number of favourable outcomes =â€‹ 8

∴

*P*(getting a prime number) = $\frac{8}{20}=\frac{2}{5}$

(iv) Out of the given numbers, the numbers divisible by 10 are 10 and 20.

Let E

_{4}be the event of getting a number not divisible by 10.

Then, number of favourable outcomes = 20 − 2 = 18â€‹

∴

*P*(getting a number not divisible by 10) = $\frac{18}{20}=\frac{9}{10}$

#### Page No 611:

#### Question 21:

15 cards numbered 1, 2, 3, 4,..., 14, 15 are put in a box and mixed thoroughly. A man draws a card at random from the box. Find the probability that the number on the card is

(i) an even number

(ii) a prime number

(iii) divisible by 3

(iv) divisible by 2 and 3

#### Answer:

The possible outcomes are 1,2, 3,4, 5................15.

Number of all possible outcomes = 15

(i) Out of the given numbers, the even numbers are 2, 4, 6, 8, 10, 12 and 14.

Let E_{1} be the event of getting an even number.

Then, number of favourable outcomes = 7â€‹

∴* P* (getting an even number) = $\frac{7}{15}$

(ii) Out of the given numbers, the prime numbers are 2, 3, 5, 7, 11 and 13.

Let E_{2} be the event of getting a prime number.

Then,â€‹ number of favourable outcomes =â€‹ 6

∴* P* (getting a prime number) = $\frac{6}{15}=\frac{2}{5}$

(iii) Out of the given numbers, the numbers divisible by 3 are 3, 6, 9, 12 and 15.

Let E_{3} be the event of getting a number divisible by 3.

∴

*P*(getting a number divisible by 3) = $\frac{5}{15}=\frac{1}{3}$

(iv) Out of the given numbers, the numbers divisible by 2 and 3 are 6 and 12.

_{4}be the event of getting a number that is divisible by 2 and 3.

Then,â€‹ number of favourable outcomes =â€‹ 2

∴

*P*(getting a number divisible by 2 and 3) = $\frac{2}{15}$

#### Page No 611:

#### Question 22:

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 E_{1} be the event of getting an even number.

Then, number of favourable outcomes = 50â€‹ $\left[{T}_{n}=100\Rightarrow 2+(n-1)\times 2=100,\Rightarrow n=50\right]\phantom{\rule{0ex}{0ex}}$

∴* 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 E_{2} be the event of getting a number less than 16.

Then,â€‹ number of favourable outcomes =â€‹ 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 E_{3} be the event of getting a number that is a perfect square.

∴

*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 E

_{4}be the event of getting a prime number less than 40.

Then,â€‹ number of favourable outcomes =â€‹ 12

∴

*P*(getting a prime number less than 40) = $\frac{12}{100}=\frac{3}{25}$

#### Page No 611:

#### Question 23:

Cards marked with numbers 13, 14, 15, ..., 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the drawn card is (i) divisible by 5 (ii) a perfect square.

#### Answer:

The possible outcomes are 13,14, 15, 16................60.

Number of all possible outcomes = 48

(i) Out of the given numbers, the numbers divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55 and 60.

Let E_{1} be the event of getting a number divisible by 5.

Then, number of favourable outcomes = 10â€‹

∴* P* (getting a number divisible by 5) = $\frac{10}{48}=\frac{5}{24}$

(ii) Out of the given numbers, the perfect squares are 16, 25, 36, and 49.

Let E_{2} be the event of getting a perfect square.

Then, number of favourable outcomes = 4â€‹

∴* P* (getting a perfect square) = $\frac{4}{48}=\frac{1}{12}$

#### Page No 611:

#### 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.

(i) Out of the given numbers, the prime numbers less than 10 are 5 and 7.

Let E

_{1}be the event of getting a prime number less than 10.

Then, number of favourable outcomes = 2â€‹

∴

*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 E

_{2}be the event of getting a perfect square.

Then, number of favourable outcomes = 5â€‹

∴

*P*(getting a perfect square) = $\frac{5}{46}$

#### Page No 611:

#### Question 25:

Find the probability of getting 53 Fridays in a leap year.

#### Answer:

A leap year has 366 days, i.e. 52 weeks and 2 days. These two days can be

(i) Sunday -Monday (ii) Monday - Tuesday,

(iii) Tuesday - Wednesday (iv) Wednesday - Thursday

(vii) Saturday - Sunday

Out of these 7 cases, 2 have Fridays.

∴

*P*(getting 53 Fridays) = $\frac{2}{7}$

#### Page No 611:

#### Question 26:

If the probability of winning a game is 0.6, what is the probability of losing it?

#### Answer:

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*) = 0.6

Now, *P*(*E*) + â€‹*P*(not *E*) = 1

⇒0.6 + â€‹*P*(not *E*) = 1

â€‹⇒â€‹*P*(not *E*) = 1− 0.6 = 0.4

∴ *P*(losing the game) = â€‹*P*(not *E*) â€‹= 0.4

#### Page No 611:

#### Question 27:

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) = â€‹$\frac{2}{52}=\frac{1}{26}$

#### Page No 612:

#### Question 28:

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

#### Page No 612:

#### Question 29:

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

#### Page No 612:

#### Question 30:

A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither a queen nor a jack.

#### Answer:

Total number of all possible outcomes= 52

There are 4 queens and 4 jacks in a pack of 52 cards.

Therefore, number of cards, each one of which is either a queen or a jack = 4 + 4 = 8

Let E be the event that the card drawn is neither a queen nor a jack.

Then, the number of favourable outcomes = (52 − 8) = 44

∴ *P*( getting neither a queen nor a jack) = *P* (*E*) = $\frac{44}{52}=\frac{11}{13}$

#### Page No 612:

#### Question 31:

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)

*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

*P*(getting a king or a queen ) = $\frac{2}{13}$

Now, â€‹

*P*(

*E*) + â€‹

*P*(not

*E*) = 1

∴ â€‹

*P*(getting neither a king nor a queen ) = $1-\left(\frac{2}{13}\right)=\frac{11}{13}$

#### Page No 612:

#### Question 32:

A card is drawn at random from a well-shuffled deck of playing cards. Find the probability of drawing a (i) face card (ii) card which is neither a king nor a red card.

#### Answer:

Total number of all possible outcomes= 52

(i) We know that kings, queens and jacks are the face cards.

_{1}be the event of getting a face card.

Then, the number of favourable outcomes = 12

∴

*P*( getting a face card) =

*P*(

*E*

_{1}) = $\frac{12}{52}=\frac{3}{13}$

(ii) There are 26 red cards (including 2 kings) and there are 2 more kings.

Therefore, number of cards, each one of which is either a red card or a king = 26 + 2 = 28

Let E

_{2}be the event that the card drawn is neither a king nor a red card.

Then, the number of favourable outcomes = (52 − 28) = 24

∴

*P*( getting neither a king nor a red card) =

*P*(

*E*

_{2}) = $\frac{24}{52}=\frac{6}{13}$

#### Page No 612:

#### Question 33:

The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card is (i) heard (ii) queen (iii) clubs.

#### Answer:

After removing the king, queen and jack of clubs, the number of remaining cards = (52 − 3) = 49

∴

*P*( getting a card of hearts) = $\frac{13}{49}$

(ii) Number of queens = 3

∴

*P*(getting a queen) = $\frac{3}{49}$

(iii) There are 10 cards of clubs in the remaining 49 cards.

∴

*P*( getting a card of clubs) = $\frac{10}{49}$

#### Page No 612:

#### Question 34:

The king, the queen, the jack and 10, all of spades, are lost from a pack of 52 playing cards. A card is drawn from the remaining well-shuffled pack. Find the probability of getting a (i) red card (ii) king (iii) black card.

#### Answer:

Remaining number of cards = (52 − 4) = 48

â€‹

∴

*P*( getting a red card) = $\frac{26}{48}=\frac{13}{24}$

(ii) Number of kings = 3

∴

*P*( getting a king) = $\frac{3}{48}=\frac{1}{16}$

(iii) There are 13 cards of clubs and 9 cards of spades in the remaining 48 cards.

*P*( getting a black card) = $\frac{22}{48}=\frac{11}{24}$

#### Page No 612:

#### Question 35:

Red kings, queens and jacks are removed from a deck of 52 playing cards and then the deck is well-shuffled. A card is drawn from the remaining cards. Find the probability of getting (i) a king (ii) a red card (iii) a spade.

#### Answer:

Remaining number of cards = (52 − 6) = 46

â€‹

∴

*P*(getting a king) = $\frac{2}{46}=\frac{1}{23}$

(ii) Number of red cards = 20

∴

*P*(getting a red card) = $\frac{20}{46}=\frac{10}{23}$

(iii) Number of spade cards = 13

*P*(getting a spade card) = $\frac{13}{46}$

#### Page No 612:

#### Question 36:

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.

#### 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) = $\frac{5}{11}$

#### Page No 612:

#### Question 37:

A bag contains 27 balls of which some are white and some are red. A ball is chosen at random. The probability of getting a red ball is $\frac{2}{3}$. Find the number of white balls.

#### Answer:

Let the number of white balls in the bag be *x*.

Then, the number of red balls = ( 27 − *x*)

Given: *P*(red ball) = $\frac{2}{3}$

∴ $\frac{27-x}{27}=\frac{2}{3}\Rightarrow \frac{27-x}{9}=\frac{2}{1}\Rightarrow 27-x=18\Rightarrow x=27-18=9$

Hence, the number of white balls in the bag is 9.

#### Page No 613:

#### Question 38:

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 E_{1} be the event that the pointer rests on 6.

Then, number of favourable outcomes = 1â€‹

∴* P* (arrow pointing at 6) = *P*(* **E*_{1}) = $\frac{1}{12}$

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

_{2}be the event of getting an even number.

Then, number of favourable outcomes = 6â€‹

∴

*P*(arrow pointing at an even number) =

*P*(

*E*

_{2}) = $\frac{6}{12}=\frac{1}{2}$

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

Let E

_{3}be the event of the arrow pointing at a prime number.

Then, number of favourable outcomes =â€‹ 5

∴

*P*(arrow pointing at a prime number) =

*P*(

*E*

_{3}) = $\frac{5}{12}$

(iv) Out of the given numbers, the numbers that are multiples of 5 are 5 and 10 only.

Let E

_{4}be the event of the arrow pointing at a multiple of 5.

∴

*P*(arrow pointing at a number that is a multiple of 5) =

*P*(

*E*

_{4}) =$\frac{2}{12}=\frac{1}{6}$

#### Page No 613:

#### Question 39:

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 numbers less than 15

(ii) a number divisible by 3 and 5.

#### Answer:

Number of possible outcomes = 18 $\left[{T}_{n}=35\Rightarrow 1+(n-1)\times 2=35\Rightarrow n=18\right]$

(i) Out of the given numbers, the prime numbers less than 15 are:

_{1}be the event of getting a prime number less than 15.

Then, number of favourable outcomes = 6â€‹

∴

*P*(getting a prime number less than 15) = $\frac{6}{18}=\frac{1}{3}$

(ii) Out of the given numbers, the numbers divisible by 3 and 5 are 15 and 30.

Let E

_{2}be the event of getting a number divisible by 3 and 5 ( such numbers will be divisible by 15). Then, number of favourable outcomes = 2â€‹

∴

*P*(getting a number divisible by 3 and 5 ) = $\frac{2}{18}=\frac{1}{9}$

#### Page No 617:

#### Question 1:

Two coins are tossed simultaneously. What is the probability of getting exactly one head?

(a) $\frac{1}{4}$

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

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

(d) $\frac{2}{3}$

#### Answer:

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

Explanation:

When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH and TT.

Total number of possible outcomes = 4

Let E be the event of getting exactly one head.

â€‹Number of favourable outcomes = 2

*P*(getting exactly 1 head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{2}{4}=\frac{1}{2}$

#### Page No 617:

#### Question 2:

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.

â€‹Number of favourable outcomes = 3

*P*(getting at most 1head) = â€‹$\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{4}$

#### Page No 617:

#### Question 3:

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.

â€‹Number of favourable outcomes = 3

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{8}$

#### Page No 617:

#### Question 4:

Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

(i) $\frac{2}{4}$

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

(iii) $\frac{5}{8}$

(iv) $\frac{7}{8}$

#### Answer:

(d) $\frac{7}{8}$

Explanation:

â€‹ When 3 coins are tossed simultaneously, all 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 at most 2 heads.

*P*(getting at most 2 heads)= $\frac{\mathrm{Number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{7}{8}$

#### Page No 617:

#### Question 5:

A die is thrown once. What is the probability of getting a number less than 3?

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

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

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

(d) $\frac{1}{4}$

#### Answer:

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

Explanation:

In a single throw of a die, all possible outcomes are:

1,2,3,4,5,6

Total number of possible outcomes = 6

Let E be the event of getting a number less than 3.

Then, the favourable outcomes are 1 and 2.

Number of favourable outcomes = 2

∴ Probability of getting a number less than 3:

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

#### Page No 617:

#### Question 6:

A die is thrown once. What is the probability of getting a number more than 4?

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

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

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

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

#### Answer:

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

Explanation:

In a single throw of a die, all possible outcomes are:

1, 2, 3, 4, 5, 6

Total number of possible outcomes = 6

Let E be the event of getting a number more than 4.

Then, the favourable outcomes are 5 and 6.

Number of favourable outcomes = 2

∴ Probability of getting a number more than 4:

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

#### Page No 617:

#### Question 7:

A die is thrown once. What is the probability of getting a prime number?

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

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

(c) $\frac{2}{3}$

(d) none of these

#### Answer:

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

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*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{6}=\frac{1}{2}$

#### Page No 617:

#### Question 8:

In a throw of a die, what is the probability of getting a 4?

(a) $\frac{2}{3}$

(b) $\frac{1}{4}$

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

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

#### Answer:

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

Explanation:

In a single throw of a die, all possible outcomes are:

1, 2, 3, 4, 5, 6

Total number of possible outcomes = 6

Let E be the event of getting a 4.

Then, the favourable outcome is 4

Number of favourable outcomes = 1

∴ Probability of getting a 4 = *P *(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{1}{6}$

#### Page No 617:

#### Question 9:

Cards bearing numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out of bag at random. What is the probability that the number on the taken out is an even number?

(a) $\frac{9}{17}$

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

(c) $\frac{5}{9}$

(d) $\frac{7}{18}$

#### Answer:

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

Explanation:

All possible outcomes are 3, 4,................20.

Number of all possible outcomes = 18

Total even numbers from 3 to 20 = 9

Number of favourable outcomes = 9

∴

*P*(getting an even number) = $\frac{9}{18}=\frac{1}{2}$

#### Page No 618:

#### Question 10:

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) $\frac{2}{5}$

â€‹

Explanation:

All possible outcomes are 6, 7, 8................15.

Number of all possible outcomes = 10

Number of favourable outcomes = 4

∴

*P*(getting a number that is less than 10) = $\frac{4}{10}=\frac{2}{5}$

#### Page No 618:

#### Question 11:

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

#### Page No 618:

#### Question 12:

Two dice are thrown simultaneously. What is the probability of getting two numbers whose sum is 10?

(a) $\frac{5}{18}$

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

(c) $\frac{1}{12}$

(d) $\frac{1}{4}$

#### Answer:

(c) $\frac{1}{12}$

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 that the sum of the numbers appearing on the top of the two dice is equal to 10.

Then the favourable outcomes are:

(4,6), (6,4), (5,5)

Number of favourable outcomes = 3

∴ *P*( getting a sum equal to 10) = $\frac{3}{36}=\frac{1}{12}$

#### Page No 618:

#### Question 13:

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:

â€‹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) = $\frac{4}{20}=\frac{1}{5}$

#### Page No 618:

#### Question 14:

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:

â€‹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) = $\frac{12}{25}$

#### Page No 618:

#### Question 15:

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 ) = $\frac{8}{24}=\frac{1}{3}$

#### Page No 618:

#### Question 16:

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) $\frac{4}{5}$

â€‹

Explanation:

Number of prizes = 6

Number of blanks = 24

Total number of tickets = 6 + 24= 30

∴ *P*(not getting a prize ) = $\frac{24}{30}=\frac{4}{5}$

#### Page No 618:

#### Question 17:

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

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 ) = $\frac{7}{9}$

#### Page No 619:

#### Question 18:

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 ) = $\frac{6}{10}=\frac{3}{5}$

#### Page No 619:

#### Question 19:

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

∴

*P*(getting a ball that is not black) = $\frac{13}{15}$

#### Page No 619:

#### Question 20:

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

∴

*P*(getting a ball that is neither black nor white) = $\frac{4}{12}=\frac{1}{3}$

#### Page No 619:

#### Question 21:

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.

#### Page No 619:

#### Question 22:

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.

#### Page No 619:

#### Question 23:

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

#### Page No 619:

#### Question 24:

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) = $\frac{4}{52}=\frac{1}{13}$

#### Page No 619:

#### Question 25:

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) = $\frac{12}{52}=\frac{3}{13}$

#### Page No 619:

#### Question 26:

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

#### Page No 620:

#### Question 27:

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) = $\frac{4}{52}=\frac{1}{13}$

#### Page No 620:

#### Question 28:

Two friends were born in 2000. What is the probability that they have the same birthday?

(a) $\frac{1}{365}$

(b) $\frac{1}{366}$

(c) $\frac{2}{365}$

(d) $\frac{1}{183}$

#### Answer:

(b) $\frac{1}{366}$

Explanation:

The year 2000 was a leap year. So, it had 366 days.

One friend's birthday may fall on any one of the 366 days and the other friend's birthday may fall on any one of the 366 days.

Total number of ways in which the birthdays of the two friends fall on particular days = (366 x 366)

The favourable outcomes for both friends to have the same birthday out of the 366 days = 366

∴ *P*( both have the same birthday) = $\frac{366}{\left(366\times 366\right)}=\frac{1}{366}$

#### Page No 620:

#### Question 29:

What is the probability that two friends have different birthdays?

(a) $\frac{1}{365}$

(b) $\frac{2}{365}$

(c) $\frac{364}{365}$

(d) $\frac{363}{365}$

#### Answer:

(c) $\frac{364}{365}$

Explanation:

One friend's birthday may fall on any one of the 365 days and the other friend's birthday may also fall on any one of the 365 days.

Total number of ways in which the birthdays of both the friends fall on particular days = (365 â¨¯ 365)

The favourable outcomes for the event that both friends have the same birthday on any one of the 365 days = 365

∴ *P*( both have the same birthday) = $\frac{365}{\left(365\times 365\right)}=\frac{1}{365}$

∴ *P*( both have different birthdays) = 1- *P*( both have the same birthdays)

= $1-\left(\frac{1}{365}\right)=\frac{364}{365}$

#### Page No 620:

#### Question 30:

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

(a) $\frac{2}{7}$

(b) $\frac{1}{7}$

(c) $\frac{7}{52}$

(d) $\frac{7}{53}$

#### Answer:

(b) $\frac{1}{7}$

Explanation:

An ordinary year has 365 days, i.e. 52 weeks and 1 day.

*P*( getting 53 Mondays ) = $\frac{1}{7}$

#### Page No 620:

#### Question 31:

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*) + â€‹*P*(not *E*) = 1

⇒ 0.4 + â€‹*P*(not *E*) = 1

â€‹⇒â€‹ *P*(not *E*) = 1− 0.4 = 0.6

∴ *P*(losing the game) = â€‹*P*(not *E*) â€‹= 0.6

#### Page No 620:

#### Question 32:

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.

#### Page No 620:

#### Question 33:

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.

#### Page No 620:

#### Question 34:

If the probability of occurrence of an event is *p* then the probability of occurrence of its complementary event is

(a) *p*

(b) *p* − 1

(c) 1 − *p*

(d)$1-\frac{7}{p}$

#### Answer:

(c) 1− *p*

â€‹Explanation:*P*(not E) = 1− *P*(E) = 1− *P*

#### Page No 620:

#### Question 35:

A student calculates that the probability of his winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has he bought?

(a) 120

(b) 240

(c) 480

(d) 760

#### Answer:

(c) 480

Explanation:

Let the number of tickets bought by the student be *x*.

Then , $\frac{x}{6000}=0.08$

Then, *x* = 6000 â¨¯ 0.08 = 480

So, he bought 480 tickets.

#### Page No 620:

#### Question 36:

Fill in the blanks:

(i) The probability of an impossible event is ........ .

(ii) The probability of a sure event ......... .

(iii) If *E* is an event and *P*(*E*) = *p *then *p*(*not* *E*) = ........ .

(iv) If *E* is a possible event and it is not a sure event then *P*(*E*) lies between ...... and ....... .

(v) *P*(*E*) + *P*(*not* *E*) = ........ .

(vi) The probability of occurrence of an event cannot be greater than ....... .

#### Answer:

(i) 0

(ii) 1

(iii) 1−* p*

(iv) 0, 1

(v) 1

(vi)1

#### Page No 621:

#### Question 37:

Match the following columns:

Column I | Column II |

(a) In tossing a fair die, the probability of getting an odd number or a number less than 4 is ....... . |
(p) $\frac{1}{23}$ |

(b) A card is drawn from a well-shuffled deck of 52 cards. The probability of getting a black face card is ........ . | (q) $\frac{5}{36}$ |

(c) Two dice are thrown simultaneously. The probability of getting a total of 8 is ....... . | (r) $\frac{3}{26}$ |

(d) Red kings, queens and jacks are removed from a deck of 52 cards. A card is drawn from the remaining cards. The probability of getting a king is ........ . | (s) $\frac{2}{3}$ |

#### Answer:

(a) - (s), (b) - (r), (c) - (q), (d) - (p)

Explanation:

(a) 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 an odd number or a number less than 4.

Then, the odd numbers are 1, 3 and 5 and the numbers less than 4 are 1, 2 and 3.

Therefore, number of favourable outcomes = [ 1, 2, 3, 5] = 4

∴ *P*( getting an odd number or a number less than 4) = *P *(*E*) = $\frac{4}{6}=\frac{2}{3}$

(b) Total number of all possible outcomes= 52

We know that kings, queens and jacks are face cards.

Then, the number of favourable outcomes = 6

∴

*P*(getting a black face card) =

*P*(

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

(c) Number of all possible outcomes = 36

Let

*E*be the event that the sum of the numbers appearing on the top of the two dice is equal to 8.

Then the favourable outcomes are:

(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

Number of favourable outcomes = 5

∴

*P*(getting a sum equal to 8) = $\frac{5}{36}$

(d) We know that there are 6 red face cards.

Total number of all possible outcomes= (52 − 6) = 46

Now, number of kings left in the deck = 2

Let

*E*be the event of getting a king.

Then, the number of favourable outcomes = 2

∴

*P*( getting a king) =

*P*(

*E*) = $\frac{2}{46}=\frac{1}{23}$

#### Page No 622:

#### Question 38:

**Assertion (A)**

If the probability of winning a game is $\frac{8}{15}$, then the probability of losing the game is $\frac{7}{15}$.**Reason (R)**

For any event *E*, we have P(E) + *P*(not* E*) = 1.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

#### Answer:

â€‹â€‹(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

â€‹Explanation:

Let *E* be the event of winning a game.

Then, (not *E*) is the event of not winning the game or losing the game.

Then, *P*(*E*) = $\frac{8}{15}$

Now, *P*(*E*) + â€‹*P*(not *E*) = 1 ⇒ $\frac{8}{15}$ + â€‹*P*(not *E*) = 1

â€‹⇒â€‹*P*(not *E*) = $1-\left(\frac{8}{15}\right)=\frac{7}{15}$

∴ *P*(losing the game) = â€‹*P*(not *E*) â€‹= â€‹$\frac{7}{15}$

#### Page No 622:

#### Question 39:

**Assertion (A)**

Cards numbered 1, 2, 3, ..., 15 are put in a box and mixed thoroughly. One card is then drawn at random. The probability of drawing an even number is $\frac{1}{2}$.**Reason (R)**

For any event *E,* we have:

0 ≤ *P*(*E*) ≥ 1.

#### Answer:

(d) Assertion (A) is false and Reason (R) is true.

Explanation:

The expression 0 ≤ *P*(*E*) ≤ 1 is clearly true.

∴ Reason ( R) is true.

Total number of cards: 1,2,3................15

Number of all possible outcomes = 15

Out of these, the even numbers are 2, 4, 6, 8, 10, 12 and 14.

Total even numbers from 1 to 15 = 7

Number of favourable outcomes = 7

∴* P* (drawing an even number) = $\frac{7}{15}\ne \frac{1}{2}$

∴ Assertion (A) is false.

Thus, Assertion (A) is false and Reason ( R) is true.

So, the correct answer is (d).

#### Page No 626:

#### Question 1:

A bag contains 3 red balls, 5 white balls and 7 black balls. A ball is drawn at random. What is the probability that the ball drawn is neither red nor black?

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

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

(c) $\frac{7}{15}$

(d) $\frac{8}{15}$

#### Answer:

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

Explanation:

Total possible outcomes = Total number of balls = ( 3 + 5 + 7 ) = 15

*P*(getting a ball that is neither red nor black) = $\frac{5}{15}=\frac{1}{3}$

#### Page No 626:

#### Question 2:

If an event cannot occur then its probability is

(a) 1

(b) −1

(c) 0

(d) none of these

#### Answer:

(c) 0

For an impossible event, the probability is 0.

#### Page No 627:

#### Question 3:

Two dice are thrown at random. What is the probability that the sum of numbers obtained will be 9?

(a) $\frac{1}{36}$

(b) $\frac{1}{18}$

(c) $\frac{2}{9}$

(d) $\frac{1}{9}$

#### Answer:

(d) $\frac{1}{9}$

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 two numbers whose sum is 9.

Then, the favourable outcomes are:

(3, 6), (4, 5) , (5, 4) (6, 3)

Number of favourable outcomes = 4

∴ *P*(getting two numbers whose sum is 9 ) = *P* ( *E*) = $\frac{4}{36}=\frac{1}{9}$

#### Page No 627:

#### Question 4:

If *P*(*E*) = 0.5, then *P*(*not E*) = ?

(a) 0.05

(b) 0.95

(c) $\frac{1}{0.05}$

(d) $\frac{1}{0.95}$

#### Answer:

(b) 0.95

Explanation: *P*(*E*) + â€‹*P*( not *E*) = 1

∴ â€‹*P*( not *E*) = 1 â€‹− *P*(*E*)

⇒ 1 − 0.05 = 0.95

#### Page No 627:

#### Question 5:

A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice the probability of drawing a red ball, find the number of blue balls.

#### Answer:

Let the number of blue balls in the bag be *x*.

Then, total number of balls = ( 6 + *x*)

Given: *P*(blue ball) = 2 â¨¯ *P*( red ball)

∴ $\frac{x}{6+x}=2\times \left(\frac{6}{6+x}\right)\Rightarrow x=12$

Hence, the number of blue balls in the bag is 12.

#### Page No 627:

#### Question 6:

An integer is chosen from 1 to 50. What is the probability that it is divisible by 8?

#### Answer:

The possible outcomes are 1, 2, 3, 4, 5................50.

Number of all possible outcomes = 50

Number of favourable outcomes = 6

∴

*P*(getting a number that is divisible by 8) = $\frac{6}{50}=\frac{3}{25}$

#### Page No 627:

#### Question 7:

Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from the box. Find the probability that the number on the card is a perfect square.

#### Answer:

The possible outcomes are 2, 3, 4, 5................101.

Number of all possible outcomes = 100

Number of favourable outcomes = 9

∴

*P*(getting a number that is a perfect square) = $\frac{9}{100}$

#### Page No 627:

#### Question 8:

A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is not defective?

#### Answer:

Total number of possible outcomes = 20

Number of defective bulbs = 4

∴ Number of non-defective bulbs = 20 − 4 = 16

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

Then, number of favourable outcomes = 16

∴ *P*(getting a non-defective bulb) = *P*(*E*) = $\frac{16}{20}=\frac{4}{5}$

#### Page No 627:

#### Question 9:

A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is neither an ace nor a king.

#### Answer:

Total number of all possible outcomes= 52

Let *E* be the event of getting either a king or an ace.

Then, ( not *E*) is the event that the drawn card is neither a king nor an ace.

Number of aces = 4

Favourable outcomes for event

*E*= 4 + 4 = 8

Favourable outcomes for event (not

*E)*= 52 − 8 = 44

*P*(getting neither a king nor an ace ) = $\frac{44}{52}=\frac{11}{13}$

#### Page No 627:

#### Question 10:

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 *E*_{1} be the event of getting A.

Number of favourable outcomes = 3

∴ *P*( getting A) = *P *(*E*_{1}) = $\frac{3}{6}=\frac{1}{2}$

(ii) Let *E*_{2} be the event of getting B.

Number of favourable outcomes = 2

∴ *P*( getting B) = *P *(*E*_{2}) = $\frac{2}{6}=\frac{1}{3}$

#### Page No 627:

#### Question 11:

A box contains 30 cards, numbered 1 to 30. A card is drawn from the box at random. Find the probability that the number of the drawn card is (i) even, (ii) prime and (iii) multiple of 7.

#### Answer:

(i) Out of the given numbers, the even numbers are :

_{1}be the event of getting an even number.

Then, number of favourable outcomes = 15â€‹

∴

*P*(getting an even number) = $\frac{15}{30}=\frac{1}{2}$

(ii) Out of the given numbers, the prime numbers are:

_{2}be the event of getting a prime number.

Then, number of favourable outcomes = 10â€‹

∴

*P*(getting a prime number) = $\frac{10}{30}=\frac{1}{3}$

(iii) Out of the given numbers, the multiples of 7 are:

7, 14, 21 and 28

Let E

_{3}be the event of getting a multiple of 7.

Then, number of favourable outcomes = 4â€‹

∴

*P*(getting a multiple of 7) = $\frac{4}{30}=\frac{2}{15}$

#### Page No 627:

#### Question 12:

A bag contains 4 white balls, 5 red balls, 8 black balls and 3 blue balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is (i) red or blue (ii) not black and (iii) neither black nor red.

#### Answer:

Total possible outcomes = Total number of balls

*P*(getting a red or blue ball ) = $\frac{8}{20}=\frac{2}{5}$

(ii) Total number of balls that are not of black colour = (4 + 5 + 3) = 12

∴

*P*(getting a ball that is not of black colour ) = $\frac{12}{20}=\frac{3}{5}$

(iii ) Total number of balls that are not black or red = (4 +3) = 7

∴

*P*(getting a ball that is neither black nor red) = $\frac{7}{20}$

#### Page No 627:

#### Question 13:

A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability that the card drawn is (a) a king or a jack (b) a red card (c) neither a king nor a queen.

#### Answer:

Total number of all possible outcomes= 52

(a) Total number of kings = 4

Total number of jacks = 4

Let *E* be the event of getting either a king or a jack.

Then, the favourable outcomes = 4 + 4 = 8

∴ *P*( getting a king or a jack) = *P* (*E*) = $\frac{8}{52}=\frac{2}{13}$

(b) Number of red cards = 26

∴ *P*( getting a red card) = $\frac{26}{52}=\frac{1}{2}$

(c) Let *E* be the event of getting either a king or a queen. Then, (not *E*) is the event that the drawn card is neither a king nor a queen.

Total number of queens = 4

Then, the favourable outcomes for event

*E*= 4 + 4 = 8

∴

*P*( getting a king or a queen) =

*P*(

*E*) = $\frac{8}{52}=\frac{2}{13}$

*P*(

*E*) + â€‹

*P*(not

*E*) = 1

∴ â€‹

*P*(getting neither a king nor a queen ) =

*P*(not

*E*) = $1-\left(\frac{2}{13}\right)=\frac{11}{13}$

#### Page No 627:

#### Question 14:

Find the probability that a number selected at random from the numbers 1, 2, 3, ..., 34 , 35 is (a) a prime number, (b) a multiple of 7, (c) a multiple of 3 or 5.

#### Answer:

The possible outcomes are 1, 2, 3, 4, 5................35.

Number of all possible outcomes = 35

(a) Out of the given numbers, the prime numbers are

_{1}be the event of getting a prime number.

Then,â€‹ number of favourable outcomes =â€‹ 11

∴

*P*(getting a prime number) =

*P*(

*E*

_{1}) = $\frac{11}{35}$

(b) Out of the given numbers, the numbers that are multiples of 7 are

_{2}be the event of getting a multiple of 7.

∴

*P*(getting a multiple of 7 ) =

*P*(

*E*

_{2}) $\frac{5}{35}=\frac{1}{7}$

(c) Out of the given numbers, the numbers that are multiples of 3 are

And the multiples of 5 are 5, 10, 15, 20, 25, 30 and 35.

_{3}be the event of getting a multiple of 3 or 5.

∴

*P*(getting a multiple of 3 or 5 ) =

*P*(

*E*

_{3}) = $\frac{16}{35}$

#### Page No 628:

#### Question 15:

A jar contains 24 marbles out of which some 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 in the jar = 24

Let the number of green marbles in the jar be *x*.

Then, number of blue marbles = ( 24 − *x*)

Given: *P*(green marble) =$\frac{2}{3}$

Therefore,

$\frac{x}{24}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=48\phantom{\rule{0ex}{0ex}}\Rightarrow x=16\phantom{\rule{0ex}{0ex}}$

Thus, there are 16 green marbles in the jar and number of blue marbles = 24 − 16 = 8

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

#### Page No 628:

#### Question 16:

A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card drawn is (a) card of spades or an ace (b) a face card.

#### Answer:

Total number of all possible outcomes= 52

(i) Number of spade cards = 13

Number of aces = 4 (including 1 of spade)

*P*( getting a spade or an ace card) = $\frac{16}{52}=\frac{4}{13}$

(ii) Number of face cards = ( 4 kings + 4 queens + 4 jacks) = 12

∴

*P*( getting a face card) = $\frac{12}{52}=\frac{3}{13}$

#### Page No 628:

#### Question 17:

A bag contains tickets numbered 11, 12, 13, 14, ..., 30. A ticket is taken out form the bag at random. Find the probability that the number on the drawn ticket is (a) a multiple of 7 (b) greater than 15 and a multiple of 5.

#### Answer:

â€‹Total number of tickets = 20

(i) Out of the given ticket numbers, the multiples of 7 are = 14, 21 and 28

_{1}be the event of getting a multiple of 7.

∴

*P*(getting a multiple of 7 ) = $\frac{3}{20}$

(ii) Out of the given ticket numbers, the multiples of 5 are = 15, 20, 25 and 30

_{2}be the event of getting a number that is greater than 15 and a multiple of 5.

∴

*P*(getting a ticket number that is greater than 15 and a multiple of 5 ) = $\frac{3}{20}$

#### Page No 628:

#### Question 18:

From a pack of 52 cards, jacks, queens, kings and aces of red colour are removed. From the remaining cards, one card is drawn at random. Find the probability that the card drawn is (a) a black queen (b) a picture card.

#### Answer:

After removing jacks, queens, kings, and aces of red colour, the number of remaining cards = 52 − ( 2 + 2 + 2 + 2) = 44

∴

*P*( getting a black queen) = $\frac{2}{44}=\frac{1}{22}$

(ii) Number of picture cards left among the remaining 44 cards = ( 2 kings + 2 queens + 2 jacks) = 6

∴

*P*( getting a picture card) = $\frac{6}{44}=\frac{3}{22}$

#### Page No 628:

#### Question 19:

A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at ran dom from the bag. What is the probability that the ball drawn is (a) white or blue, (b) not white, (c) neither white nor black?

#### Answer:

Total possible outcomes = Total number of balls

*P*(getting a white or blue ball ) = $\frac{7}{18}$

(b) Total number of balls that are not of white colour = (7 + 4 + 2) = 13

∴

*P*(getting a ball that is not of white colour ) = $\frac{13}{18}$

(c ) Total number of balls that are neither white nor black = (7 + 2) = 9

∴

*P*(getting a ball that is neither white nor black) = $\frac{9}{18}=\frac{1}{2}$

#### Page No 628:

#### 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 2 more queens are there.

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

View NCERT Solutions for all chapters of Class 10