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The probability that a man will be alive in $20$ years is $\frac{3}{5}$ and the probability that his wife will be alive in $20$ years is $\frac{2}{3}$. Then the probability that at least one will be alive in $20$ years, is
$\frac{{13}}{{15}}$
$\frac{7}{{15}}$
$\frac{4}{{15}}$
None of these
Solution
(a) Let $A$ be the event that the husband will be alive $20$ years.
$B$ be the event that the wife will be alive $20$ years.
Clearly $A$ and $B$ are independent events
$\therefore$ $P(A \cap B) = P(A).\,P(B)$
Given $P(A) = \frac{3}{5},$ $P(B) = \frac{2}{3}$
The probability that at least of them will be alive $20$ years is
$P(A \cup B) = P(A) + P(B) – P(A \cap B)$
$ = P(A) + P(B) – P(A)P(B) $
$= \frac{3}{5} + \frac{2}{3} – \frac{3}{5}.\frac{2}{3} = \frac{{9 + 10 – 6}}{{15}} = \frac{{13}}{{15}}$
Aliter : Required probability is $1 – P(A$ and $B$ both will die)
$ = 1 – \frac{2}{5} \times \frac{1}{3} = 1 – \frac{2}{{15}} = \frac{{13}}{{15}}.$
