# RD Sharma Solutions for Class 10 Maths Chapter 13 Probability

## RD Sharma Solutions Class 10 Maths Chapter 13 – Free PDF Download

RD Sharma Solutions for Class 10 Maths Chapter 13 – Probability are provided here. These solutions are collated by subject matter experts to boost in-depth knowledge of concepts among students. For better academic performance, students of Class 10 are recommended to follow RD Sharma Solutions on a daily basis. This also helps them to enhance their time management and problem-solving skills, which are crucial for the board exam.

For more conceptual knowledge, students can refer to RD Sharma Solutions for Class 10 Maths Chapter 13, created by the expert faculty at BYJU’S. The solutions are explained with pictorial representation for better understanding among students. RD Sharma Solutions for Class 10 Maths offer descriptive answers in a simple and lucid manner in accordance with students’ understanding capacity. Students can download the chapter solutions updated for the 2023-24 syllabus from the links given below.

### RD Sharma Class 10 Chapter 13 Exercise 13.1 Page No: 13.20

1. The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?

Solution:

Given: Probability that it will rain tomorrow P(E) = 0.85

Required to find: Probability that it will not rain tomorrow P(E).

We know that sum of the probability of occurrence of an event and the probability of non-occurrence of an event is 1.

Therefore, the probability that it will not rain tomorrow is = 0.15

2. A die is thrown. Find the probability of getting:

(i) a prime number   (ii) 2 or 4

(iii) a multiple of 2 or 3   (iv) an even prime number

(v) a number greater than 5           (vi) a number lying between 2 and 6

Solution:

Given: A dice is thrown once.

Required to find:

(i) Probability of getting a prime number.

(ii) Probability of getting 2 or 4.

(iii) Probability of getting a multiple of 2 or 3.

(iv) Probability of getting an even number.

(v) Probability of getting a number greater than five.

(vi) Probability of lying between 2 and 6.

The total number on a dice is 6, i.e., 1, 2, 3, 4, 5 and 6.

(i) Prime numbers on dice are 2, 3, and 5. So, the total number of prime numbers is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a prime number = 3/6 = 1/2

(ii) For getting 2 and 4, clearly, the number of favourable outcomes is 2.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting 2 or 4 = 2/6 = 1/3

(iii) Multiple of 2 are 3 are 2, 3, 4 and 6.

So, the number of favourable outcomes is 4.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a multiple of 2 or 3 = 4/6 = 2/3

(iv) An even prime number is 2 only.

So, the number of favourable outcomes is 1.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an even prime number = 1/6

(v) A number greater than 5 is 6 only.

So, the number of favourable outcomes is 1.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a number greater than 5 = 1/6

(vi) Total number on a dice is 6.

Numbers lying between 2 and 6 are 3, 4 and 5

So, the total number of numbers lying between 2 and 6 is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a number lying between 2 and 6 = 3/6 =1/2

3. Three coins are tossed together. Find the probability of getting:

(i) exactly two heads (ii) at most two heads

(iii) at least one head and one tail (iv) no tails

Solution:

Given: Three coins are tossed simultaneously.

When three coins are tossed, then the outcome will be any one of these combinations.

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

So, the total number of outcomes is 8.

(i) For exactly two heads, the favourable outcome are THH, HHT and HTH.

So, the total number of favourable outcomes is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting exactly two heads is 3/8.

(ii) For getting at least two heads, the favourable outcomes are HHT, HTH, HHH, and THH.

So, the total number of favourable outcomes is 4.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting at least two heads when three coins are tossed simultaneously = 4/8 = 1/2

(iii) For getting at least one head and one tail, the cases are THT, TTH, THH, HTT, HHT and HTH.

So, the total number of favourable outcomes, i.e., at least one tail and one head, is 6.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting at least one head and one tail = 6/8 = 3/4

(iv) For getting an outcome of no tail, the only possibility is HHH.

So, the total number of favourable outcomes is 1.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting no tails is 1/8.

4. A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number

Solution:

Given: A pair of dice is thrown.

Required to find: Probability that the total of numbers on the dice is greater than 9.

First, let’s write the all possible events that can occur.

(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),

It’s seen that the total number of events is 62 = 36

Favourable events, i.e., getting the total of numbers on the dice greater than 9, are

(5,5), (5,6), (6,4), (4,6), (6,5) and (6,6).

So, the total number of favourable events, i.e., getting the total of numbers on the dice greater than 9, is 6

We know that, Probability = = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting the total of numbers on the dice greater than 9 = 6/36 = 1/6

5. Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.

Solution:

Given: A pair of dice is thrown.

Required to find: Probability that the total of numbers on the dice is greater than 10.

First, let’s write the all possible events that can occur.

(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),

It’s seen that the total number of events is 62 = 36

Favourable events, i.e., getting the total of numbers on the dice greater than 10, are (5, 6), (6, 5) and (6, 6).

So, the total number of favourable events, i.e., getting the total of numbers on the dice greater than 10, is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting the total of numbers on the dice greater than 10 = 3/36 = 1/12

6. A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:

(i) a black king

(ii) either a black card or a king

(iii) black and a king

(iv) a jack, queen or a king

(v) neither a heart nor a king

(vi) spade or an ace

(vii) neither an ace nor a king

(viii) neither a red card nor a queen

(ix) the seven of clubs

(x) a ten

(xii) a black card

(xiii) a seven of clubs

(xiv) jack

(xv) the ace of spades

(xvi) a queen

(xvii) a heart

(xviii) a red card

Solution:

Given: A card is drawn at random from a pack of 52 cards.

Required to Find: Probability of the following.

The total number of cards in a pack = 52

(i) Number of cards which are black king = 2

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a black king = 2/52 = 1/26

(ii) The otal number of black cards is (13 + 13) 26

The total number of kings is 4, of which 2 black kings are also included.

So, the total number of black cards or kings will be 26+2 = 28

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a black card or a king = 28/52 = 7/13

(iii) The total number of cards which are a black card and a king card, is 2.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a black card and a king is 2/52 = 1/26

(iv) A jack, queen, or king are 3 from every 4 suits.

So, the total number of a jack, queen and king is 12.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a jack, queen or king is 12/52 = 3/13

(v) The total number of heart cards is 13, and kings are 4, in which the king of heart is also included.

So, the total number of cards that are a heart and a king = 13 + 3 = 16

Hence, the total number of cards that are neither a heart nor a king = 52 – 16 = 36

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards, neither a heart nor a king = 36/52 = 9/13

(vi) The total number of spade cards is 13.

The total number of aces is 4, in which the ace of a spade is included in the number of spade cards.

Hence, the total number of cards which is a spade or ace = 13 + 3 = 16

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards that are a spade or an ace = 16/52 = 4/13

(vii) The total number of ace cards is 4 and the king is 4.

The total number of cards that are an ace or a king = 4 + 4 = 8

So, the total number of cards that are neither an ace nor a king is 52 – 8 = 44

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards which are neither an ace nor a king = 44/52 = 11/13

(viii) It’s known that the total number of red cards is 26.

The total number of queens is 4, in which 2 red queens are also included,

Hence, the total number of red cards or queens will be 26 + 2 = 28

So, the total number of cards that are neither a red nor a queen= 52 -28 = 24

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting neither a red card nor a queen = 24/52 = 6/13

(ix) The total number of cards other than ace is 52 – 4 = 48

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting other than ace = 48/52 = 12/13

(x) The total number of tens in the pack of cards is 4.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a ten = 4/52 = 1/13

(xi) The total number of spades is 13.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a spade = 13/52 = 1/4

(xii) The total number of black cards in the pack is 26.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting black cards is 26/52 = 1/2

(xiii) The total number of 7 of the club is 1 only.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a 7 of club = 1/52

(xiv) The total number of jacks is 4.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a jack = 4/52 = 1/13

(xv) The total number of the ace of a spade is 1.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an ace of spade = 1/52

(xvi) The total number of queens is 4.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a queen = 4/52 = 1/13

(xvii) The total number of heart cards is 13.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a heart card = 13/52 = 1/4

(xviii) The total number of red cards is 26.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a red card = 26/52 = 1/2

(xix) The total number of kings and queens is 4 + 4 = 8

So, the total number of cards that are neither a king nor a queen is 52 – 8 = 44

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a card which is neither a queen nor a king = 44/52 = 11/13

7. In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability that the drawn ticket bears a prime number.

Solution:

Given: Tickets are marked numbers from 1 to 50. And one ticket is drawn at random.

Required to find: Probability of getting a prime number on the drawn ticket.

The total number of tickets is 50.

Tickets which are numbered as prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

The total number of tickets marked as prime is 15.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a prime number on the ticket = 15/50 = 3/10

8. An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.

Solution:

Given: A bag contains 10 red and 8 white balls.

Required to find: Probability that one ball is drawn at random and gets a white ball.

The otal number of balls 10 + 8 = 18

The total number of white balls is 8.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a white ball from the urn is 8/18 = 4/9

9. A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:

(i) white? (ii) red?

(iii) black? (iv) not red

Solution:

Given: A bag contains 3 red, 5 black and 4 white balls.

Required to find: Probability of getting a

(i) White ball

(ii) Red ball

(iii) Black ball

(iv) Not red ball

The total number of balls 3 + 5 + 4 =12

(i) The total number of white balls is 4.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a white ball = 4/12 = 1/3

(ii) The total number of red balls is 3.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a red ball = 3/12 = 1/4

(iii) The total number of black balls is 5.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a black ball = 5/12

(iv) The total number of balls which are not red is 4 white balls and 5 black balls, i.e., 4 + 5 = 9

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting no red ball = 9/12 = 3/4

10. What is the probability that a number selected from the numbers 1, 2, 3, …, 15 is a multiple of 4?

Solution:

Given: Numbers are from 1 to 15. One number is selected,

Required to find: Probability that the selected number is a multiple of 4,

The total number is between 1 to 15 to 15.

Numbers that are multiples of 4 are 4, 8 and 12.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of selecting a number which a multiple of 4 is 3/15 = 1/5

11. A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that the ball drawn is not black?

Solution:

Given: A bag contains 6 red, 8 black and 4 white balls and a ball is drawn at random.

Required to find: Probability that the ball drawn is not black.

The total number of balls 6 + 8 + 4 = 18

The total number of black balls is 8.

So, the total number of balls which are not black is 18 – 8 = 10

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a ball which is not black = 10/18 = 5/9

12. A bag contains 5 white balls and 7 red balls. One ball is drawn at random. What is the probability that the ball drawn is white?

Solution:

Given: A bag contains 7 red and 5 white balls, and a ball is drawn at random.

Required to find: Probability that the ball drawn is white.

The total number of balls 7 + 5 = 12

The total number of white balls is 5.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a white ball = 5/12

13. Tickets numbered from 1 to 20 are mixed up, and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7?

Solution:

Given: Tickets marked from 1 to 20 are mixed up. One ticket is picked at random.

Required to find: Probability that the ticket bears a multiple of 3 or 7.

The total number of cards is 20.

And the cards marked which are multiple of 3 or 7 are 3, 6, 7, 9, 12, 14, 15 and 18.

So, the total number of cards marked multiple of 3 or 7 is 8.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a card that is a multiple of 3 or 7 is 8/20 = 2/5

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

Solution:

Given: In a lottery, there are 10 prizes and 25 blanks.

Required to find: Probability of winning a prize.

The total number of tickets is 10 + 25 = 35

The total number of prize-carrying tickets is 10.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of winning a prize = 10/35 = 2/7

15. If the probability of winning a game is 0.3, what is the probability of losing it?

Solution:

Given: Probability of winning a game P(E) = 0.3

To Find: Probability of losing the game.

We know that the sum of the probability of occurrence of an event and the probability of non-occurrence of an event is 1.

So,

Thus, the probability of losing the game is

$$\begin{array}{l}P(\bar{E}) = 0.7\end{array}$$

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

(i) red (ii) black or white (iii) not black

Solution:

Given: A bag contains 7 red, 5 black and 3 white balls and a ball is drawn at random.

Required to find: Probability of getting a

(i) Red ball

(ii) Black or white ball

(iii) Not a black ball

The total number of balls 7 + 5 + 3 = 15

(i) The total number of red balls is 7.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a red ball = 7/15

(ii) Total number of black or white balls is 5 + 3 = 8

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a white or black ball = 8/15

(iii) The total number of black balls is 5.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a black ball P(E) = 5/15 = 1/3

But this is not what is required.

And, we know that the sum of the probability of occurrence of an event and the probability of non-occurrence of an event is 1.

Thus, the probability of drawing a card that is not black is 2/3.

17. A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

(i) White (ii) Red

(iii) Not black (iv) Red or White

Solution:

Given: A bag contains 4 red, 5 black and 6 white balls, and a ball is drawn at random.

Required to Find: Probability of getting a

(i) White ball

(ii) Red ball

(iii) Not a black ball

(iv) Red or white

Total number of balls 4 + 5 + 6 = 15

(i) The total number of white balls in the bag is 6.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a white ball = 6/15 = 2/5

(ii) The total number of red balls in the bag is 4.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a red ball = 4/15

(iii) The total number of black balls is 5.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a black ball P(E) = 5/15 = 1/3

We know that the sum of the probability of occurrence of an event and the probability of non-occurrence of an event is 1.

Thus, the probability of drawing a ball that is not black is 2/3.

(iv) The total number of red or white balls 4 + 6 = 10

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a white or red ball = 10/15 = 2/3

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

(i) A king of the red suit (ii) A face card (iii) A red face card

(iv) A queen of the black suit (v) A jack of hearts (vi) A spade

Solution:

Given: One card is drawn from a well-shuffled deck of 52 playing cards.

Required to find: Probability of the following:

The total number of cards is 52.

(i) The total number of cards which is the king of the red suit is 2.

The number of favourable outcomes, i.e., the total number of cards which is the king of the red suit, is 2.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards which is the king of the red suit = 2/52 = 1/26

(ii) The total number of face cards is 12.

The number of favourable outcomes, i.e., the total number of face cards, is 12.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a face card = 12/52 = 3/13

(iii) The total number of red face cards is 6.

The number of favourable outcomes, i.e., the total number of red face cards, is 6.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a red face card = 6/52 = 3/26

(iv) The total number of the queen of black suit cards is 2.

The total number of favourable outcomes, i.e., the total number of the queen of black suit cards, is 2.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards which is a queen of black suit = 2/52 = 1/26

(v) The total number of jack of hearts is 1.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a card which is a jack of hearts = 1/52

(vi) The total number of spade cards is 13.

The total number of favourable outcomes, i.e., the total number of spade cards, is 13.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a spade card = 13/52 = 1/4

19. Five cards – ten, jack, queen, king, and an ace of diamonds are shuffled face downwards. One card is picked at random.

(i) What is the probability that the card is a queen?

(ii) If a king is drawn first and put aside, what is the probability that the second card picked up is the (a) ace? (b) king?

Solution:

Given: Five cards-ten, jack, queen, king and Ace of diamonds are shuffled face downwards.

Required to find: Probability of the following:

The total number of cards is 5.

(i) The total number of cards which is a queen is 1.

Number of favourable outcomes, i.e., the total number of cards which is queen = 1

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting cards which is a queen = 1/5

(ii) If a king is drawn first and put aside, then

The total number of cards becomes 4.

(a) The number of favourable outcomes, i.e., the total number of the ace card, is 1.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an ace card = 1/4

(b) The number of favourable outcomes, i.e., the total number of king cards, is 0.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a king = 0

20. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:

(i) Red (ii) Back

Solution:

Given: A bag contains 3 red and 5 black balls. A ball is drawn at random.

Required to find: Probability of getting a

(i) red ball

(ii) white ball

The total number of balls 3 + 5 = 8

(i) The total number of red balls is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a red ball = 3/8

(ii) The total number of the black ball is 5.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of drawing a black ball = 5/8

21. 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) 10? (ii) an odd number?

(iii) a number which is multiple of 3? (iv) an even number?

Solution:

Given: A game of chance consists of spinning an arrow which is equally likely to come to rest pointing numbers 1, 2, 3 ….12

Required to find: Probability of following:

The total numbers on the spin are 12.

(i) Favourable outcomes, i.e., to get 10 is 1.

So, the total number of favourable outcomes, i.e., to get 10, is 1.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a 10 = 1/12

(ii) Favourable outcomes, i.e., to get an odd number, are 1, 3, 5, 7, 9, and 11.

So, the total number of favourable outcomes, i.e., to get a prime number is 6.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a prime number = 6/12 = 1/2

(iii) Favourable outcomes, i.e., to get a multiple of 3, are 3, 6, 9, and 12.

So, the total number of favourable outcomes, i.e., to get a multiple of 3, is 4.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a multiple of 3 = 4/12 = 1/3

(iv) Favourable outcomes, i.e., to get an even number, are 2, 4, 6, 8, 10, and 12.

So, the total number of favourable outcomes, i.e., to get an even number is 6.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an even number = 6/12 = 1/2

22. In a class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for the class monitor. What she does is she writes the name of each pupil on a card and puts them into a basket, and mixes them thoroughly. A child is asked to pick one card from the basket. What is the probability that the name written on the card is:

(i) The name of a girl (ii) The name of a boy?

Solution:

Given: In a class, there are 18 girls and 16 boys, and the class teacher wants to choose one name. The class teacher writes all pupils’ names on a card and puts them in the basket, and mixes them well thoroughly. A child picks one card.

Required to find: The probability that the name is written on the card is

(i) The name of a girl

(ii) The name of a boy

The total number of students in the class = 18 + 16 = 34

(i) The names of a girl are 18, so the number of favourable cases is 18.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting the name of a girl on the card = 18/34 = 9/17

(ii) The names of a boy are 16, so the number of favourable cases is 16.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting the name of a boy on the card = 16/34 = 8/17

23. Why is tossing a coin considered to be a fair way of deciding which team should choose ends in a game of cricket?

Solution:

No. of possible outcomes while tossing a coin = 2, i.e., 1 head or 1 tail.

Probability = Number of favourable outcomes/ Total number of outcomes

P(getting head) = 1/2

P(getting tail) = 1/2

As we can see that the probability of both events is equal, these are called equally like events.

Thus, tossing a coin is considered to be a fair way of deciding which team should choose ends in a game of cricket.

24. What is the probability that a number selected at random from the numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 will be their average?

Solution:

Given numbers are 1, 2, 2, 3, 3, 3, 4, 4, 4, 4.

The total number of possible outcomes = 10

= 30/10

= 3

Now, let E be the event of getting 3.

The number of favourable outcomes = 3 {3, 3, 3}

P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 3/10

Therefore, the probability that a number selected at random will be the average is 3/10.

25. There are 30 cards, of the same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 3.

Solution:

Given: 30 cards of the same size in a bag on which numbers 1 to 30 are written. And one card is taken out of the bag at random.

Required to find: Probability that the number on the selected card is not divisible by 3.

The total number of possible outcomes is 30 {1, 2, 3, … 30}.

Let E = The event of getting a number that is divisible by 3

So, the number of favourable outcomes = 10{3, 6, 9, 12, 15, 18, 21, 24, 27, 30}

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 10/30

= 1/3

Then, = Event of getting number not divisible by 3

= 2/3

Thus, the probability that the number on the selected card is not divisible by 3 = 2/3

26. A bag contains 5 red, 8 white and 7 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.

Solution:

The total number of possible outcomes = 20 (5 red, 8 white & 7 black}

(i)  Let E = The event of drawing a red or white ball

No. of favourable outcomes = 13 (5 red + 8 white)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 13/20

(ii) Let E = The event of getting a black ball

No. of favourable outcomes =7 (7 black balls)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 7/20

= Event of not getting a black ball

(iii) Let E = The event of getting neither a white nor a black ball

No. of favourable outcomes = 20 – 8 – 7 = 5(total balls – no. of white balls – no. of black balls)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 5/20 = 1/4

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

Solution:

The total no. of possible outcomes = 25 {1, 2, 3…. 25}

Let E = Event of getting a prime no.

So, the favourable outcomes are 2, 3, 5, 7, 11, 13, 17, 19, 23

No. of favourable outcomes = 9

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 9/25

The,
= Event of not getting a prime

Therefore, the probability of selecting a number which is not prime is 16/25.

28. A bag contains 8 red, 6 white and 4 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

Solution:

The total number of balls = 8 + 6 + 4 = 18

The total no. of possible outcomes =18

(i)  Let E = The event of getting a red or white ball

No. of favourable outcomes = 14 (8 red balls + 6 white balls)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 14/18

P(E) = 7/9

(ii)  Let E = The event of getting a black ball

Number of favourable outcomes = 4 (4 black balls)

P(E) = 4/18

P(E) = 2/9

Then,
= Event of not getting a black ball

(iii)  Let E = The event of getting neither a white nor a black ball

No. of favourable outcomes = 18 – 6 – 4

= 8(Total balls – no. of white balls – no. of black balls)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 8/18 = 4/9

29. Find the probability that a number selected at random from the numbers 1, 2, 3…. 35 is a:

(i) Prime number (ii) Multiple of 7 (iii) Multiple of 3 or 5

Solution:

Numbers 1, 2, 3….. 35 are a total of 35.

The total no. of possible outcomes = 35

(i) Let E = The event of getting a prime number

No. of favourable outcomes =11 {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 11/35

(ii) Let E = The event of getting a number which is a multiple of 7

No. of favourable outcomes = 5 {7, 14, 21, 28, 35}

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 5/35 = 1/7

(iii)  Let E = The event of getting a no. which is a multiple of 3 or 5

No. of favourable outcomes = 16 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 5, 10, 20, 25, 35}

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 16/35

30. From a pack of 52 playing cards, Jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is

(i) a black queen (ii) a red card

(iii) a black jack (iv) a picture card (Jacks, queens and kings are picture cards.)

Solution:

We know that,

The total no. of cards = 52

All Jacks, queens, kings and aces of red colour are removed.

The total no. of possible outcomes = 52 – 2 – 2 – 2 – 2 = 44 (remaining cards)

(i) Let E = The event of getting a black queen

No. of favourable outcomes = 2 (Queen of spade and club)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 1/22

(ii)  Let E = The event of getting a red card

No. of favourable outcomes = 26 – 8

= 18 (total red cards Jacks, queens, kings and aces of red colour)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 18/44 = 9/22

(iii)  Let E = The event of getting a black Jack

No. of favourable outcomes = 2 (jack of club & spade)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 2/44 = 1/22

(iv) Let E = The event of getting a picture card

No. of favourable outcomes = 6 (2 Jacks, 2 kings and 2 queens of black colour)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 6/44 = 3/22

31. A bag contains lemon-flavoured candies only. Malini 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

Solution:

(i)  We know that the bag contains lemon-flavoured candies only. So, the event that Malini takes out an orange-flavoured candy is an impossible event.

Thus, the probability of an impossible event is 0.

P(an orange flavoured candy) = 0

(ii)   The bag contains lemon-flavoured candies only. Then, the event that Malini will take out a lemon-flavoured candy is a sure event. Thus, the probability of a sure event is 1.

P(a lemon flavoured candy) = 1

32. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution:

If E = The event of 2 students not having the same birthday

Given, P(E) = 0.992

Let,
= The event of 2 students having the same birthday

We know that,

Therefore, the probability that the 2 students have the same birthday is 0.008.

33. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red (ii) not red

Solution:

Given,

A bag contains 3 red and 5 black balls.

So, the total no. of possible outcomes = 8 (3 red + 5 black)

(i) Let E = The event of getting a red ball.

No. of favourable outcomes = 3 (as there are 3 red)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 3/8

(ii)  Let
= The event of getting no red ball.

From the previous question, we already have P(E) = 3/8

34. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red (ii) not green

Solution:

Given,

A box containing 5 red, 8 white and 4 green marbles.

So, the total no. of possible outcomes = 17 (5 red + 8 white + 4 green)

(i) Let E = The event of getting a red marble

Number of favourable outcomes = 5 (as 5 red marbles)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 5/17

(ii) Let E= The event of getting a green marble

Number of favourable outcomes = 4 (as 4 green marbles)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 4/17

So,
= The event of getting not a green marble

Then, we know that

Therefore, the probability that the marble is taken out is not green is 13/17.

35. A lot consists of 144 ball pens, of which 20 are defective and others good. Nuri will buy a pen if it is good but will not buy 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.

Solution:

No. of good pens = 144 – 20 = 124

No. of detective pens = 20

The total no. of possible outcomes =144 (total no. of pens)

(i) For her to buy it, the pen should be a good one.

So, let E = The event of buying a pen which is good.

No. of favourable outcomes = 124 (124 good pens)

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 124/144 = 31/36

(ii)  Now, Let
= The event of her not buying a pen as it was a defective one.

Therefore, the probability that she will not buy = 5/36

36. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at the pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Solution:

We have,

No. of good pens = 132

No. of defective pens = 12

So, the total no. of pens = 132 + 12 = 144

Then, the total no. of possible outcomes = 144

Now, let E = The event of getting a good pen.

No. of favourable outcomes = 132 {132 good pens}

Probability, P(E) = Number of favourable outcomes/ Total number of outcomes

P(E) = 132/144 = 11/12

### RD Sharma Class 10 Chapter 13 Exercise 13.2 Page No: 13.32

1. Suppose you drop a tie at random on the rectangular region shown in fig. below. What is the probability that it will land inside the circle with diameter 1 m?

Solution:

Area of a circle with radius 0.5 m A circle = (0.5)2 = 0.25 πm2

Area of rectangle = 3 x 2 = 6m2

The probability that the tie will land inside the circle with diameter 1m is

Therefore, the probability that the tie will land inside the circle = π/24

2. In the accompanying diagram, a fair spinner is placed at the centre O of the circle. Diameter AOB and radius OC divide the circle into three regions labelled X, Y and Z.? If ∠BOC = 45°. What is the probability that the spinner will land in region X?

Solution:

Given,

∠BOC = 45°

∠AOC = 180 – 45 = 135° [Linear Pair]

Area of circle = πr2

Area of region x = θ/360 × πr2

= 135/360 × πr2

= 3/8 × πr2

The probability that the spinner will land in the region is

x = 3/8

Therefore, the probability that the spinner will land in region X is 3/8.

3. A target, shown in fig. below, consists of three concentric circles of radii – 3, 7 and 9 cm, respectively. A dart is thrown and lands on the target. What is the probability that the dart will land in the shaded region?

Solution:

We have,

1st circle – with radius 3

2nd circle – with radius 7

3rd circle – with radius 9

So, their areas would be

Area of 1st circle = π(3)2 = 9π

Area of 2nd circle = π(7)2 = 49π

Area of 3rd circle = π(9)2 = 81π

Area of shaded region = Area of 2nd circle – Area of 1st circle

= 49π − 9π

= 40π

The probability that it will land in the shaded region is

Therefore, the probability that the dart will land in the shaded region is 40/81.

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