# RS Aggarwal Solutions Chapter 15 Probability Exercise 15B Class 10 Maths

 Chapter Name RS Aggarwal Chapter 15 Probability Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 15AExercise 15C Related Study NCERT Solutions for Class 10 Maths

### Exercise 15B Solutions

1. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card was drawn is

(i) a card of a spade or an Ace

(ii) a red king

(iii) either a king or queen

(iv) neither a king nor the queen.

Solution

Total number of all possible outcomes = 52

(i) Number of space cards = 13

Number of aces = 4 (including 1 of spade)

Therefore, number of spade cards and aces = (13 + 4 – 1) = 16

Therefore, P(getting a spade or an ace card) = 16/52 = 4/13

(ii) Number of red kings = 2

Therefore, P(getting a red king) = 2/52 = 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 favorable outcomes = 4 + 4 = 8

Therefore, P(getting a king or a queen) = P(E) = 8/52 = 2/13

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

Then, P(getting a king or a queen) = 2/13

Now, P(E) + P(not E) = 1

Therefore, P(getting neither a king nor a queen) = 1 – 2/13 = 11/13

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

(i) divisible by 2 or 3,

(ii) a prime number.

Solution

Total number of outcomes = 25

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

Out of given numbers, numbers divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24.

Out of the given numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21 and 24.

Out of the given numbers, numbers divisible by both 2 and 3 are 6, 12, 18 and 24.

Number of favorable outcomes = 16

Therefore, P(getting a card divisible by 2 or 3) = P(E1) = (Number of outcomes favorable to E1)/(Number of all possible outcomes) = 16/25

Thus, the probability that the number on the drawn card is divisible by 2 or 3 is 16/25.

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

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

Number of favorable outcomes = 9

Therefore, P(getting a prime number) = P(E2) = (Number of outcomes favorable to E2)/(Number of all possible outcomes) = 9/25

Thus, the probability that the number on the drawn card is a prime number is 9/25.

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

Solution

Given numbers 3, 5, 7, 9, ….., 35, 37 from an AP with a = 3 and d = 2.

Let Tn = 37. Then,

3 + (n – 1)2 = 37

⇒ 3 + 2n – 2 = 37

⇒ 2n = 36

⇒ n = 18

Thus, total number of outcomes = 18

Let E be the event of getting a prime number.

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

The number of favorable outcome = 11.

Therefore, P(getting a prime number) = P(E) = (Number of outcomes favorable to E)/(Number of all possible outcomes) = 11/18

Thus, the probability that the number on the card is a prime number is 11/18.

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

(i) not divisible by 3

(ii) a prime number greater than 7,

(iii) not a perfect square number.

Solution

The total number of outcome = 30

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

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

Number of favorable outcomes = 30 – 10 = 20

Therefore, P(getting a number not divisible by 3) = P(E1) = (Number of outcomes favorable to E1)/(Number of all possible outcomes) = 20/30 = 2/3.

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

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

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

Number of favorable outcomes = 6

Therefore, P(getting a prime number greater than 7) = P(E2) = (Number of outcomes favorable to E2)/(Number of all possible outcomes) = 6/30 = 1/5.

Thus, the probability that the number on the card is a prime number greater than 7 is 1/5.

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

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

Number of favorable outcomes = 30 – 5 = 25

Therefore, P(getting non-perfect square number) = P(E3) = (Number of outcomes favorable to E3)/(Number of all possible outcomes) = 25/30 = 5/6

Thus, the probability that the number on the card is not a perfect square number is 5/6.

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

(i) a prime number less than 15

(ii) a number divisible by 3 and 15.

Solution

Given number 1, 3, 5, …, 35 from an AP with a = 1 and d = 2.

Let Tn = 35, Then,

1 + (n – 1)2 = 35

⇒ 1 + 2n – 2 = 35

⇒ 2n = 36

⇒ n = 18

Thus, total number of outcomes = 18.

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

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

The number of favorable outcomes = 5

Therefore, P(getting a prime number less than 15) = P(E1) = (Number of outcomes favorable to E1)/(Number of all possible outcomes) = 5/18

Thus, the probability of getting a card bearing a prime number less than 5 is 5/18.

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

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

The number of favorable outcomes = 1.

Therefore, P(getting a number divisible by 3 and 5) = P(E2) = (Number of outcomes favorable to E2)/(Number of all possible outcomes) = 1/18

Thus, the probability of getting a card bearing a number divisible by 3 and 5 is 1/18.

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

(i) a 1 digit number,

(ii) a number divisible by 5,

(iii) an odd number less than 30,

(iv) a composite number between 50 and 70.

Solution

Given numbers 6, 7, 8, ……, 70 from an AP with a = 6 and d = 1

Let Tn = 70. Then,

6 + (n – 1)1 = 70

⇒ 6 + n – 1 = 70

⇒ n = 65

Thus, total number of outcomes = 65

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

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

The number of favorable outcomes = 4.

Therefore, P(getting a one-digit number) = P(E1) = (Number of outcomes favorable to E1)/(Number of all possible outcomes) = 4/65

Thus, the probability that the card bears a one-digit number is 4/65.

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

Out of these numbers, numbers divisible by 5 are 10, 15, 20, ….., 70.

Given number 10, 15, 20, …, 70 form an AP with a = 10 and d = 5.

Let Tn = 70. Then,

10 + (n – 1)5 = 70

⇒ 10 + 5n – 5 = 70

⇒ 5n = 65

⇒ n = 13

Thus, number of favorable outcomes = 13

Therefore, P(getting a number divisible by 5) = P(E2) = (Number of outcomes favorable to E2)/(Number of all possible outcomes) = 13/65 = 1/5

Thus, the probability that the card bears a number divisible by 5 is 1/5.

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

Out of these numbers, odd numbers less than 30 are 7, 9, 11, ….., 29.

Given number 7, 9, 11, ….., 29 from an AP with a = 7 and d = 2

Let Tn = 29. Then,

7 + (n – 1)2 = 29

⇒ 7 + 2n – 2 = 29

⇒ 2n = 24

⇒ n = 12

Thus, number of favorable outcomes = 12

Therefore, P(getting a odd number less than 30) = P(E3) = (Number of outcomes favorable to E3)/(Number of all possible outcomes) = 12/65

Thus, the probability that the card bears an odd number less than 30 is 12/65.

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

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

Number of favorable outcomes = 15

Therefore, P(getting a composite number between 50 and 70) = P(E4) = (Number of outcomes favorable to E4)/(Number of all possible outcomes) = 15/65 = 3/13

Thus, probability that the card bears composite number between 50 and 70 is 3/13.

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

(i) Less than 19,

(ii) a prime number less than 20.

Solution

Given number 1, 3, 5, …, 101 from an AP with a = 1 and d = 2

Let Tn = 101. Then,

1 + (n – 1)2 = 101

⇒ 1 + 2n – 2 = 101

⇒ 2n = 102

⇒ n = 51

Thus the total number = 51.

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

Out of these numbers, numbers less than 19, are 1, 3, 5, ….., 17.

Given number 1, 3, 5, …, 17 from an AP with a = 1 and d = 2.

Let Tn = 17. Then,

1 + (n – 1)2 = 17

⇒ 1 + 2n – 2 = 17

⇒ 2n = 18

⇒ n = 9

Thus, number of favorable outcomes = 9.

Therefore, P(getting a number less than 19) = P(E1) = (Number of outcomes favorable to E1)/(Number of all possible outcomes) = 9/51 = 3/17

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

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

Thus, the number of favorable outcomes = 7

Therefore, P(getting a prime number less than 20) = P(E2) = (Number of outcomes favorable to E2)/(Number of all possible outcomes) = 7/51

Thus, the probability that the number on the drawn card is a prime number less than 20 is 7/51.

8. Tickets numbered 2, 3, 4, 5, …….., 100, 101 are placed in a box and mix 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.

Solution

All possible outcomes are 2, 3, 4, 5, ….., 101.

Number of all possible outcomes = 100

(i) Out of these the numbers that are even = 2, 4, 6, 8, ……, 100

Let E1 be the event of getting an even number.

Then, number of favorable outcomes = 50

[Tn = 100 ⇒ 2 + (n – 1) × 2 = 100, ⇒ n = 50)

Therefore, P(getting an even number) = 50/100 = 1/2

(ii) Out of these, the numbers that are less than 16 = 2, 3, 4, 5, …, 15.

Let E2 be the event of getting a number less than 16.

Then, number of favorable outcomes= 14

Therefore, P(getting a number less than 16) = 14/10 0 = 7/50

(iii) Out of these, the numbers that are perfect squares = 4, 9, 16, 25, 36, 49, 64, 81 and 100

Let E3 be the event of getting a number that is a perfect square.

Then, number of favorable outcomes = 9

Therefore, P(getting a number that is a perfect square) = 9/100

(iv) Out of these, prime numbers less than 40 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Let E4 be the event of getting a prime number less than 40.

Then, number of favorable outcomes = 12

Therefore, P(getting a prime number less than 40 = 12/100 = 3/25

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

Solution

The total number of outcomes = 80

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

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

Thus, the number of favorable outcomes = 8

Therefore, P(getting a perfect square number) = P(E1) = (Number of outcomes favorable to E1)/(Number of all possible outcomes) = 8/80 = 1/10

Thus, the probability that the disc bears a perfect square number is 1/10.

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

(i) will be a Rs 1 coin ?

(ii) will not be a Rs 5 coin

(iii) will be 50-p or a Rs 2.coin?

Solution

Number of 50-p coins = 100.

Number of Rs 1 coins = 70

Number of Rs 2 coins = 50

Number of Rs 5 coins = 30.

Thus, the total number of outcomes = 250

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

The number of favorable outcomes = 70

Therefore, P(getting a Rs 1 coin) = P(E1) = (Number of outcomes favorable of E1)/(Number of all possible outcomes) = 70/250 = 7/25

Thus, the probability that the coin will be a Rs 1 coin is 7/25.

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

Number of favorable outcomes = 250 – 30 = 220

Therefore, P(not getting a Rs. 5 coin) = P(E2) = (Number of outcomes favorable to E2)/(Number of all possible outcomes) = 220/250 = 22/25

Thus, probability that the coin will not be a Rs 5 coin is 22/25.

(iii) Let E3 be the event of getting a 50-p or a Rs. 2 coins

Number of favorable outcomes = 100 + 50 = 150

Therefore, P(not getting a 50-p or a RS. 2 coin) = P(E3) = (Number of outcomes favorable to E3)/(Number of all possible outcomes) = 150/250 = 3/5.

Thus, probability that the coin will be a 50-p or a Rs 2 coin is 3/5.

11. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is 1/4. The probability of selecting a blue ball at random from the same jar is 1/3. If the jar contains 10 orange balls, find the total number of balls in the jar.

Solution

It is given that,

P(getting a red ball) = 1/4 and P(getting a blue ball) = 1/3

Let P(getting an orange ball) be x.

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

Therefore, 1/4 + 1/3 + x = 1

⇒ x = 1 – 1/4 - 1/3

⇒ x = (12 – 3 – 4)/12

⇒ x = 5/12

Therefore, P(getting an orange ball) = 5/12.

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

Therefore, P(getting an orange ball) = 10/n

⇒ 10/n = 5/12

⇒ n = 24

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