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Question
Fatima and John appear in an interview for two vacancies for the same post. The probability of Fatima's selection is \[\frac{1}{7}\] and that of John's selection is \[\frac{1}{5}\] What is the probability that
(i) both of them will be selected?
(ii) only one of them will be selected?
(iii) none of them will be selected?
Solution
\[P\left( \text{ Fatima gets selected } \right) = P\left( A \right) = \frac{1}{7}\]
\[P\left( \text{ John gets selected } \right) = P\left( B \right) = \frac{1}{5}\]
\[\left( i \right) P\left( \text{ both of them get selected} \right) = P\left( A \cap B \right)\]
\[ = P\left( A \right) \times P\left( B \right)\]
\[ = \frac{1}{7} \times \frac{1}{5} = \frac{1}{35}\]
\[\left( ii \right) P\left( \text{ only one of them gets selected } \right) = P\left( A \right) \times P\left( \bar{B} \right) + P\left( \bar{A} \right) \times P\left( B \right)\]
\[ = \frac{1}{7}\left( 1 - \frac{1}{5} \right) + \left( 1 - \frac{1}{7} \right)\frac{1}{5}\]
\[ = \frac{1}{7} \times \frac{4}{5} + \frac{6}{7} \times \frac{1}{5}\]
\[ = \frac{4}{35} + \frac{6}{35}\]
\[ = \frac{10}{35} = \frac{2}{7}\]
\[\left( iii \right) P\left( \text{ none of them get selected } \right) = P\left( \bar{B} \right) \times P\left( \bar{A} \right)\]
\[ = \left( 1 - \frac{1}{5} \right) \times \left( 1 - \frac{1}{7} \right)\]
\[ = \frac{4}{5} \times \frac{6}{7}\]
\[ = \frac{24}{35}\]
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