# SCERT Kerala Class 10 Mathematics/ Mathematics of Chance

**SCERT Kerala Class 10 Mathematics/ Mathematics of Chance is about the extra questions for practice. Here you can find out practice problems for Class 10 Mathematics.**

**SCERT Kerala Class 10 Mathematics/ Mathematics of Chance – Chapter 3Extra Questions for Practice/ Model QuestionsAnswer the following:**

**Consider all two digit numbers.**

i) What is the probability of getting a number 12?

ii) What is the probability of getting a perfect square?

Solution:

Total numbers of two digit numbers are 90.

i) Probability (getting a number 12) = 1/90

b) Two digit perfect squares are 16, 25, 36, 49, 64 and 81.

Probability (getting a perfect square) = 6/90 = 1/15.**A box contains 4 green balls and 3 blue balls and another box contains 5 green balls and 6 blue balls.**

i) What is the probability of getting a blue ball from the first box?

ii) What is the probability of getting a green ball from the second box?

Solution:

i) Total balls in the first box = 7

Probability (getting a blue ball) = 3/7

ii) Total balls in the second bag = 11

Probability (getting a green ball) = 5/11**In a box there are 25 blue balls, 15 green balls and 10 red balls. One ball is taking at random from the box.**

i) Find the probability of getting a blue ball?

ii) Find the probability of getting a green ball or red ball?

iii) Find the probability of not getting a red ball?

Solution:

i) Total number of balls = 25 + 15 + 10 = 50 balls

Number of blue balls = 25

Probability (getting a blue ball) = 25/50 = ½

ii) Total number of green balls and red balls = 15 + 10 = 25

Probability (getting a green ball or red ball) = 25/50 = ½

iii) Probability (not getting a red ball) = 40/50 = 4/5**A box contains four slips numbered 1, 2, 3, 4 and another contains three slips numbered 1, 2, 3. If one slip is taken from each**

i) What is the probability of getting the sum 5?

ii) What is the probability of getting the product 6?

Solution:

Possible outcomes are (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4). Total 12 pairs are there.

i)Number of pairs whose sum 5 are (2, 3), (3, 2), (4, 1)

Probability (getting sum 5) = 3/12

ii) Number of pairs whose product 6 are (2, 3), (3, 2)

Probability (getting the product 6) = 2/12 = 1/6**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 boys?

ii) What is the probability of both being girls?

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

Solution:

Total number of students in 10A = 30 + 20 = 50

Total number of students in 10B = 15 + 25 = 40

Total pair of students = 50 x 40 = 2000

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

Probability (both being boys) = 450/ 2000 = 9/40

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

Probability (both being girls) = 500/2000 = ¼

iii) Number of pairs in which one is a boy and the other is a girl = 30 x 15 + 20 x 25

= 450 + 500 = 950

Probability (one being a boy and other a girl) = 950/2000 = 19/40.