The probability that a man will be alive in 2 | vedclass

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(A) Let $A$ be the event that the husband will be alive in $20$ years and $B$ be the event that the wife will be alive in $20$ years.Given $P(A) = \fr

Vedclass · Accepted Answer

$\frac{13}{15}$

Vedclass · Answer

(A) Let $A$ be the event that the husband will be alive in $20$ years and $B$ be the event that the wife will be alive in $20$ years.Given $P(A) = \frac{3}{5}$ and $P(B) = \frac{2}{3}$.Since $A$ and $B$ are independent events,the probability that at least one of them will be alive is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Since $A$ and $B$ are independent,$P(A \cap B) = P(A) 	imes P(B) = \frac{3}{5} 	imes \frac{2}{3} = \frac{6}{15} = \frac{2}{5}$.Therefore,$P(A \cup B) = \frac{3}{5} + \frac{2}{3} - \frac{6}{15} = \frac{9 + 10 - 6}{15} = \frac{13}{15}$.Alternatively,the probability that at least one is alive is $1 - P(	ext{both die}) = 1 - (1 - P(A))(1 - P(B)) = 1 - (\frac{2}{5} 	imes \frac{1}{3}) = 1 - \frac{2}{15} = \frac{13}{15}$.

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:

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