Question

A man and his wife appear for an interview for two posts. The probability of the husband’s selection is $\dfrac{1}{7}$ and that of the wife’s selection is $\dfrac{1}{5}$ .What is the probability that only one of them will be selected.
(A) $\dfrac{1}{7}$
(B) $\dfrac{2}{7}$
(C)$\dfrac{3}{7}$
(D) None of these.

Answer 1

In the question the given is probability of wife to be selected and husband to be selected. Here, we have to choose the correct probability that only one of them will be selected. By using the given and probability relation we have to find the probability of the required event.

Formula used: The probability formula is used to compute the probability of an event to occur.
Probability formula for addition,
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
For independent events A & B,
$P(A \cap B) = P\left( A \right).P\left( B \right)$.

Complete step-by-step answer:
It is given that the man and his wife appear for an interview for two posts. The probability of the husband’s selection is $\dfrac{1}{7}$ and that of the wife’s selection is $\dfrac{1}{5}$.
We need to find out the probability that only one of them will be selected.
Let, A= Event that the husband is selected.
B= event that wife is selected.
Given that, $P(A) = \dfrac{1}{7}$, $P(B) = \dfrac{1}{5}$.
The probability that husband is not selected for the post=$P(\overline A ) = 1 - \dfrac{1}{7} = \dfrac{6}{7}$ .
The probability that wife is not selected for the post=$P(\overline B ) = 1 - \dfrac{1}{5} = \dfrac{4}{5}$ .
The probability that only one of them will be selected=
=P[(A and not B)or(Band not A)]
$\Rightarrow P\left[ {\left( {A \cap \overline B } \right)or\left( {B \cap \overline A } \right)} \right] = P\left( {A \cap \overline B } \right) + P\left( {B \cap \overline A } \right)$
Rewrite the relation we get,
$\Rightarrow P\left( A \right).P\left( {\overline B } \right) + P\left( B \right).P\left( {\overline A } \right)$
Substituting the probability values,
$\Rightarrow \left( {\dfrac{1}{7} \times \dfrac{4}{5}} \right) + \left( {\dfrac{1}{5} \times \dfrac{6}{7}} \right)$
Simplifying we get,
$\Rightarrow \dfrac{4}{{35}} + \dfrac{6}{{35}}$
Adding the terms we get,
$\Rightarrow \dfrac{{10}}{{35}}$
$\Rightarrow \dfrac{2}{7}$
$\therefore$ Hence, the probability that only one of the husband or wife will be selected is $\dfrac{2}{7}$.

So, the correct answer is “Option B”.

Note: When a random experiment is entertained, one of the first questions that come in our mind is: What is the probability that a certain event occurs? A probability is a chance of prediction. When we assume that, let’s say, x be the chances of happening an event then at the same time $1 - x$ are the chances for “not happening” of an event.