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A man and his wife appear for an interview for two posts. The probability of husband’s selection is \( \frac{1}{7} \) and that of the wife’s selection is \( \frac{1}{5} \) . Find the probability that only one of them will be selected.

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Solution :
Let A be the event that husband is selected and B be the event that the wife is selected
then $P(A) =\large\frac{1}{7}$ and $P(B) =\large\frac{1}{5}$
$P(\bar {A}) =\bigg(1- \large\frac{1}{7} \bigg) =\large\frac{6}{7}$
$P(\bar {B}) =\bigg(1- \large\frac{1}{5} \bigg) =\large\frac{4}{5}$
Required probability is
$P(A) .P(\bar {B}) +P( \bar {A}) . P(B) $
$\qquad = \large\frac{1}{7} \times \large\frac{4}{5} +\large\frac{1}{5} \times \large\frac{6}{7}$
$\qquad= \large\frac{2}{7}$
answered Dec 5, 2016 by meena.p

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