# Class 10 RD Sharma Solutions – Chapter 13 Probability – Exercise 13.1 | Set 2

**Question 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?**

**Solution:**

Total cards = 5

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free classeswhich will definitely help them in making a wise career choice in the future.Total queen = 1

Number of favorable outcomes = 1

Probability = Number of favorable outcomes/ Total number of outcomes

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

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

**(a) ace?**

**Solution:**

Total cards after king = 4

Number of favorable outcomes = 1

Probability = Number of favorable outcomes/ Total number of outcomes

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

**(b) king?**

**Solution:**

Number of favorable outcomes = 0

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting a king = 0

**Question 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**

**Solution:**

Total number of balls = 3 + 5 = 8

Total red balls = 3

Probability = Number of favorable outcomes/ Total number of outcomes

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

**(ii) Back**

**Solution:**

Total black ball = 5

Probability = Number of favorable outcomes/ Total number of outcomes

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

**Question 21. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number, 1, 2, 3, …., 12 as shown in figure. What is the probability that it will point to:**

**(i) 10?**

**Solution:**

Total numbers on the spin = 12

Favorable outcomes = 1

Probability = Number of favorable outcomes/ Total number of outcomes

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

**(ii) an odd number?**

**Solution:**

Favorable outcomes are 1, 3, 5, 7, 9, and 11

Favorable outcomes = 6

Probability = Number of favorable outcomes/ Total number of outcomes

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

**(iii) a number which is multiple of 3?**

**Solution:**

Favorable outcomes are 3, 6, 9, and 12

Favorable outcomes = 4

Probability = Number of favorable outcomes/ Total number of outcomes

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

**(iv) an even number?**

**Solution:**

Favorable outcomes are 2, 4, 6, 8, 10, and 12

Favorable outcomes = 6

Probability = Number of favorable outcomes/ Total number of outcomes

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

**Question 22. In a class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for class monitor. What she does, she writes the name of each pupil on a card and puts them into a basket and mixes 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**

**Solution:**

Total number of students in the class = 18 + 16 = 34

Favorable cases = 18

Probability = Number of favorable outcomes/ Total number of outcomes

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

**(ii) The name of a boy?**

**Solution:**

Favorable cases = 16

Probability = Number of favorable outcomes/ Total number of outcomes

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

**Question 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:**

Possible outcomes while tossing a coin = 2 (1 head or 1 tail)

Probability = Number of favorable outcomes/ Total number of outcomes

P(getting head) = 1/2

P(getting tail) = 1/2

As the probability of both the events are equal

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

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

**Solution:**

Total number of possible outcomes = 10

Average of the outcomes = (1 + 2 + 2+3+3+3+4+4+4+4) / 10

= 30/10

= 3

Let E be the event of getting 3.

Number of favorable outcomes = 3

P(E) = Number of favorable 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.

**Question 25. There are 30 cards, of 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:**

Total number of possible outcomes are 30 {1, 2, 3, … 30}

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

Number of favorable outcomes = 10{3, 6, 9, 12, 15, 18, 21, 24, 27, 30}

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

P(E) = 10/30

= 1/3

P(not E) = 1- 1/3

= 2/3

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

**Question 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 **

**Solution:**

(i) red or whiteTotal number of possible outcomes = 20 (5 red, 8 white & 7 black}

Number of favorable outcomes = 13 (5 red + 8 white)

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of red or white ball = 13/20

(ii) not blackNo. of favorable outcomes =7 (7 black balls)

Probability = Number of favorable outcomes/ Total number of outcomes

Probability of black ball = 7/20

= 7/20

P (not E) = 1-7/20

= 13/20

(iii) neither white nor black.Number of favorable outcomes = 20 – 8 – 7 = 5(total balls – no. of white balls – no. of black balls)

Probability = Number of favorable outcomes/ Total number of outcomes

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

**Question 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:**

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

Favorable outcomes are 2, 3, 5, 7, 11, 13, 17, 19, 23

Number. of favorable outcomes = 9

Probability= Number of favorable outcomes/ Total number of outcomes

P(E) = 9/25

P(not E) = 1-9/25

= 16/25

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

**Question 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**

**Solution:**

Total number of balls = 8 + 6 + 4 = 18

Total outcomes =18

Favorable outcomes = 14 (8 red balls + 6 white balls)

Probability = Number of favorable outcomes/ Total number of outcomes

= 14/18

= 7/9

**(ii) Not black**

**Solution:**

Let E be event of getting a black ball

Number of favorable outcomes = 4

P(E) = 4/18

P(E) = 2/9

P(not E) = 1-2/9

= 7/9

Therefore, probability of not a black ball is 7/9

**(iii) Neither white nor black**

**Solution:**

Let E be event of getting neither a white nor a black ball

Favorable outcomes = 18 – 6 – 4

= 8

Probability= Number of favorable outcomes/ Total number of outcomes

P(E) = 8/18 = 4/9

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

**(i) Prime number**

**Solution:**

Total outcomes = 35

Favorable outcomes =11 {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31)

Probability = Number of favorable outcomes/ Total number of outcomes

= 11/35

Therefore, the probability of prime number = 1/22

**(ii) Multiple of 7**

**Solution:**

Favorable outcomes = 5 {7, 14, 21, 28, 35}

Probability = Number of favorable outcomes/ Total number of outcomes

= 5/35

= 1/7

Therefore, the probability of multiple of 7 = 1/22

**(iii) Multiple of 3 or 5**

**Solution:**

Favorable outcomes = 16 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 5, 10, 20, 25, 35}

Probability = Number of favorable outcomes/ Total number of outcomes

= 16/35

Therefore, the probability of multiple of 3 or 5= 1/22

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

**(i) a black queen**

**Solution:**

Total cards = 52

All jacks, queens & kings, ace of red color are removed.

Total outcomes = 52 – 2 – 2 – 2 – 2 = 44 (remaining cards)

Favorable outcomes = 2 (queen of spade & club)

Probability = Number of favorable outcomes/ Total number of outcomes

= 2/44

=1/22

Therefore, the probability of black queen = 1/22

**(ii) a red card**

**Solution:**

Favorable outcomes = 26 – 8

= 18

Probability = Number of favorable outcomes/ Total number of outcomes

= 18/44

= 9/22

Therefore, the probability of red card = 1/22

**(iii) a black jack**

**Solution:**

Favorable outcomes = 2 (jack of club & spade)

Probability = Number of favorable outcomes/ Total number of outcomes

= 2/44

= 1/22

Therefore, the probability of black jack = 1/22

**(iv) a picture card (Jacks, queens and kings are picture cards)**

**Solution:**

Favorable outcomes = 6 (2 jacks, 2 kings & 2 queens of black color)

Probability = Number of favorable outcomes/ Total number of outcomes

= 6/44

= 3/22

Therefore, the probability of picture card = 3/22

**Question 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**

**Solution:**

Taking out orange flavored candy is an impossible event because all lemon candies are there.

Therefore, probability = 0

**(ii) a lemon flavored candy**

**Solution:**

As the bag contains all lemon candies, taking out lemon candy is sure event

Therefore, probability = 1

**Question 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:**

E = event of 2 students not having same birthday

P(E) = 0.992

P(not E) = 1 – 0.992

= 0.008

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

**Question 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**

**Solution:**

Total outcomes = 8

Favorable outcomes = 3

Probability = Number of favorable outcomes/ Total number of outcomes

= 3/8

Therefore, the probability of red ball = 3/8

**(ii) not red**

**Solution:**

P(not red) = 1- P(red)

= 1-3/8

=5/8

Therefore, the probability of not red ball = 5/8

**Question 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**

**Solution:**

Total outcomes = 17

Favorable outcomes = 5

Probability = Number of favorable outcomes/ Total number of outcomes

= 5/17

Therefore, the probability of red marble = 5/17

**(ii) not green**

Favorable outcomes = 4

Probability = Number of favorable outcomes/ Total number of outcomes

= 4/17

P(not green) = 1 – 4/17

= 13/17

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

**Question 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**

**Solution:**

Good pens = 144 – 20 = 124

Defective pens = 20

Total outcomes =144

Favorable outcomes = 124

Probability = Number of favorable outcomes/ Total number of outcomes

= 124/144

= 31/36

Therefore, the probability that she will buy = 31/36

**(ii) She will not buy it**

**Solution:**

P(buying it) = 31/36

P(not buying it) = 1-31/36

=5/36

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

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

**Solution:**

Good pens = 132

Defective pens = 12

Total outcomes = 144

Favorable outcomes = 132

Probability = Number of favorable outcomes/ Total number of outcomes

= 132/144

= 11/12

Therefore, probability of good pen = 11/12