Question

A man and his wife appear in an interview for two vacancies in the same post. The probability of husband's selection is $\dfrac{1}{7}$ and the probability of wife's selection is $\dfrac{1}{5}$. What is the assurance that only one of them is selected?

Answer 1

Here we have to find the assurance that only one of them is selected for that initially we will find the probability of both the events not happening and using the following formula we get the probability of assurance that one of them is selected.

Formula used: $\;P\left( C \right){\text{ }} = {\text{ }}P\left( {A{\text{ }} \& {\text{ }} not{\text{ }}B} \right){\text{ }} + {\text{ }}P\left( {B{\text{ }} \& {\text{ }}not{\text{ }}A} \right)$

Complete step-by-step answer:
It is given that the probability of husband being selected is $\dfrac{1}{7}$
And the probability of wife being selected is $\dfrac{1}{5}$
Let us consider A = the event that the wife is selected.
Let us consider B= the event that the husband is selected.
Then from given, we have,
$P\left( A \right) = \dfrac{1}{5}$
$P\left( B \right) = \dfrac{1}{7}$
Let C= the event that only one of them is selected.
The event that only one of them is selected can be interpreted as;

1. The wife is selected and the husband is not.
2. The husband is selected and the wife is not.
   Mathematically, statement 1 is equivalent to
   $P\left( {A{\text{ }} \& {\text{ }} not{\rm{ }}B} \right)$
   And statement 2 is equivalent to
   $P\left( {B{\text{ }} \& {\text{ }} not {\text{ }}A} \right)$ .
   Then, $\;P\left( C \right){\text{ }} = {\text{ }}P\left( {A{\text{ }}\& {\text{ }} not {\text{ }}B} \right){\text{ }} + {\text{ }}P\left( {B{\text{ }}\& {\text{ }}not{\text{ }}A} \right)$
   Since the probability function can be split we get,
   $P\left( C \right) = {\rm{ }}P\left( A \right)P\left( {not{\rm{ }}B} \right){\rm{ }} + {\rm{ }}P\left( B \right){\rm{ }}P\left( {not{\rm{ }}A} \right)$ ……… (1)
   It is clear that the events are independent.
   P\left( {not{\rm{ }}B} \right){\rm{ }} = {\rm{ }}1 - {\rm{ }}P\left( B \right){\rm{ }} = $$$${\rm{1}} - \dfrac{1}{7}= $\dfrac{6}{7}$
   P\left( {not{\rm{ }}A} \right){\rm{ }} = {\rm{ }}1 - {\rm{ }}P\left( A \right){\rm{ }} = $$$${\rm{ 1}} - \dfrac{1}{5}= $\dfrac{4}{5}$
   Substituting the values of the probabilities in (1) we get,
   $P\left( C \right) = {\rm{ }}\dfrac{1}{5}{\rm{ }} \times {\rm{ }}\dfrac{6}{7}{\rm{ }} + {\rm{ }}\dfrac{1}{7}{\rm{ }} \times \dfrac{4}{5}$
   On solving the above equation we get,
   $P(C) = \dfrac{{10}}{{35}} = \dfrac{2}{7}$
   Hence, the probability of assurance that only one of them is selected is $\dfrac{2}{7}$.

Additional Information: In probability, the set of outcomes from an experiment is known as an Event. So say for example you conduct an experiment by tossing a coin. The outcome of this experiment is the coin landing ‘heads’ or ‘tails’. These can be said to be the events connected with the experiment. So when the coin lands tails, an event can be said to have occurred.

Note: In the problem we will find the probability of the event that does not happen. That is if probability of the event happening is given we will find the probability of not happening by subtracting the probability of event happening from total probability.