# SCERT Kerala Class 10 Mathematics/ Mathematics of Chance.

**SCERT Kerala Class 10 Mathematics/ Mathematics of Chance is about the textbook solutions of Kerala Board Mathematics Textbook. Here you can find out practice problems of Class 10 Maths Chapter 3 -Mathematics of Chance .**

**SCERT Kerala Class 10 Mathematics Textbook SolutionsMathematics of Chance – Chapter 3Answer the following:**

- In class 10A, there are 30 boys and 20 girls. In 10B, there are 15 boys and 25 girls. One student is to be selected from each class.

i) What is the probability of both being girls?

ii) What is the probability of both being boys?

iii) What is the probability of one boy and one girl?

iv) What is the probability of at least one boy?

Solution:

In class 10A, total number of students = 30 + 20 = 50

In class 10B, total number of students = 15 + 25 = 40

So there are 50 x 40 = 2000 possible pairs.

i) Number of pairs in which both are girls = 20 x 25 = 500

Probability of both being girls = 500/2000 = ¼

ii) Number of pairs in which both are boys = 30 x 15 = 450

Probability of both being boys = 450/2000 = 9/40

iii) Number of pairs in which one is boy and one is girl = 2000 – (500 + 450) = 2000 – 950 = 1050

Probability of one being one boy and one girl = 1050/2000 = 21/40

iv) Number of pairs in which at least one boy = 1050 + 450 = 1500

Probability of at least one boy = 1500/2000 = ¾ - One is asked to say a two-digit number.

i) What is the probability of both digits being the same?

ii) What is the probability of the first digit being larger?

iii) What is the probability of the first digit being smaller?

Solution:

i) If both digits being the same, possible two digit numbers are

11, 22, 33, 44, 55, 66, 77, 88, 99

These are the 9 possible ways and total numbers of two digit numbers are 90.

Probability of both digits being the same = 9/90 = 1/10

ii) Two digit numbers in which first digit being larger are

10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97, 98.

These are the 45 possible ways.

Probability of the first digit being larger = 45/90 = ½

iii) Two digit numbers in which the first digit being smaller are:

12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89.

These are the 36 possible ways.

Probability of the first digit being smaller = 36/90 = 2/5 - Each two-digit number is written on a paper slip and these are all put in a box. What is the probability that the product of the digits of a number drawn is a prime number? What if three-digit numbers are used instead?

Solution:

The product of the digits of a number drawn is a prime number are

12, 13, 15, 17, 21, 31, 51, 71

These are the 8 possible two-digit numbers.

Total numbers of two-digit numbers are 90

Probability that the product of the digits of a number drawn is a prime number

= 8/90 = 4/45

The possible 3 digit numbers are 112, 113, 115, 117, 121, 131, 151, 171

Total numbers of three-digit numbers are 900.

Probability that the product of the digits of a number drawn is a prime number

= 8/900 = 4/450 = 2/225 - Two dice with faces numbered from 1 to 6 are rolled together. What are the possible sums? Which of these sums has the maximum probability?

Solution:

If two dice with faces numbered from 1 to 6 are rolled together, the possible ways 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)

Possible sums are

2, 3, 4, 5, 6, 7

3, 4, 5, 6, 7, 8

4, 5, 6, 7, 8, 9

5, 6, 7, 8, 9, 10

6, 7, 8, 9, 10, 11

7, 8, 9, 10, 11, 12

As 7 occurs more number of times (6 times), it has the most probability.

Total numbers of outcomes are 36.

Probability = 6/36 = 1/6