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 7xC1 = 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 = \(\frac{3}{13}\)