Question

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 find the probability that at least one will be alive in $20$ years.

Answer 1

Let $A$ be the event that the husband will be alive in $20$ years.

$B$ be the event that the wife will be alive in $20$ years. Clearly $A$ and $B$ are independent events.

$\therefore P(A\cap B)=P\left(A\right)P\left(B\right).$

Given:$P\left(A\right)=\frac{3}{5},P\left(B\right)=\frac{2}{3}$

The probability that at least one of them will be alive in $20$ years is

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)$

$=\frac{3}{5}+\frac{2}{3}-\frac{3}{5}.\frac{2}{3}$

$=\frac{3}{5}+\frac{2}{3}-\frac{2}{5}$

$=\frac{1}{5}+\frac{2}{3}$

$=\frac{3+10}{15}=\frac{13}{15}$