Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 16 Probability Ex 16(b) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Exercise 16(b)

Question 1.

A school has six classes 1, 2, 3, 4, 5, and 6. Classes 2, 3, 4, 5, and 6 each have the same number of students, but there is twice this number in class 1. If a student is selected at random from the school, what is the probability that he(she) will be in

Solution:

A school has six classes 1, 2, 3, 4, 5, and 6. Classes 2, 3, 4, 5, and 6 each have the same number of students, but there is twice this number in class 1.

Let the number of students in class 2, 3, 4, 5, and 6 is x each and the number of students in class 1 is 2x.

∴ The total number of students = 7x.

A student can be chosen from 7x students in ^{7x}C_{1} = 7x ways.

∴ |S| = 7x.

(i) class 1

Solution:

Probability that the student belongs to class 1, is \(\frac{2 x}{7 x}=\frac{2}{7}\)

(ii) class 2

Solution:

Probability that the student belongs to class 2, is \(\frac{x}{7 x}=\frac{1}{7}\)

Question 2.

Let a die be weighed in such a way that the probability of getting a number n is proportional to n.

Solution:

Let a die be weighed in such a way that the probability of getting a number n is proportional to n.

Let the constant of proportionality be k.

∴ P(n) = nk so that P(1) = k

P(2) = 2k, P(3) = 3k,…. P(6) = 6k

∴ P(1) + P(2) +…….+ P(6) = 1

or, k + 2k +……..+ 6k = 1

or, 21 k = 1 or, k = \(\frac{1}{21}\)

(i) Find the probability of each elementary event.

Solution:

P(1) = \(\frac{1}{21}\), P(2) = \(\frac{2}{21}\), P(3) = \(\frac{3}{21}\), P(4) = \(\frac{4}{21}\), P(5) = \(\frac{5}{21}\) , P(6) = \(\frac{6}{21}\)

(ii) Find the probability of getting an even number in a single roll of the die.

Solution:

= P(2) + P(4) + P(6)

= \(\frac{2}{21}+\frac{4}{21}+\frac{6}{21}=\frac{12}{21}\)

(iii) Find the probability of getting a prime number in a single roll of the die.

Solution:

= P(1) + P(3) + P(5)

= \(\frac{1}{21}+\frac{3}{21}+\frac{5}{21}=\frac{9}{21}\)

(iv) Find the probability of getting a prime number in a single roll of a die.

Solution:

Probability of getting a prime number = P(3) + P(5) + P(2)

= \(\frac{3}{21}+\frac{5}{21}+\frac{2}{21}=\frac{10}{21}\)

Question 3.

Five boys and three girls are playing in a chess tournament. All boys have the same probability p of winning the tournament and all the girls have the same probability q of winning. If p = 2q, find the probability that

(i) a boy wins the tournament.

(ii) a girl wins the tournament.

Solution:

5 boys and 3 girls are playing a chess tournament. All boys have the same probability P of winning the tournament and all the girls have the

same probability q of winning.

We have P(B) =p, P(G) = q.

As there are 5 boys and 3 girls,

we have 5p + 3q = 1

Now putting p = 2q,

we have 10q + 3q = 1

or, q = \(\frac{1}{13}\) ∴ p = 2q = \(\frac{2}{13}\)

∴ P(B) = 5p = \(\frac{10}{13}\),

∴ P(G) = 3p = \(\frac{3}{13}\)