Question

Two horses are considered for a race. The probability of selection of the first horse is 1/4 and that of the second is 1/3. What is the probability that :(a) both of them will be selected. (b) only one of them will be selected. (c) none of them will be selected.

Answer 1

(1) 1/12

(2) 5/12

(3) 1/2

Answer 2

Given:

The probability of selection of the first horse=1/4

The probability of selection of the second horse=1/3

To find :

(a) The probability that both of them will be selected

(b) The probability that only one of them will be selected

(c) The probability that none of them will be selected

Solution:

Let

The probability of selection of the first horse, P(A)=1/4

The probability of selection of the second horse, P(B)=1/3

P( not A)=1-P(A)=1-1/4=$\frac{4-1}{4}=\frac{3}{4}$

P(not B)=1- P(B)=1-1/3=$\frac{3-1}{3}=\frac{2}{3}$

(a)The probability that both of them will be selected

=$P(A\cap B)=P(A)\times P(B)=\frac{1}{4}\times \frac{1}{3}=\frac{1}{12}$

=$\frac{1}{12}$

(b)

The probability that only one of them will be selected

=P(A)P(not B)+P(B)P(not A)

=$\frac{1}{4}\times \frac{2}{3}+\frac{1}{3}\times \frac{3}{4}$

=$\frac{2}{12}+\frac{3}{12}$

=$\frac{2+3}{12}=\frac{5}{12}$

(c)

The probability that none of them will be selected

=P(not A)P(not B)

$=\frac{3}{4}\times \frac{2}{3}=\frac{1}{2}$

=$\frac{1}{2}$