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 the probability that at least one will be alive in 20 years is

• A. $\frac { 13 } { 15 }$
• B. $\frac { 7 } { 15 }$
• C. $\frac { 4 } { 15 }$
• D. None of these

Answer 1

Answer: (A)

$\frac { 13 } { 15 }$

Answer 2

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.

\ $P ( A \cap B ) = P ( A ) P ( B )$ .

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

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

is $P ( A \cup B ) = P ( A ) + P ( B ) - P ( A \cap B ) = P ( A ) + P ( B ) - P ( A ) \cdot P ( B )$

$= \frac { 3 } { 5 } + \frac { 2 } { 3 } - \frac { 3 } { 5 } \cdot \frac { 2 } { 3 } = \frac { 9 + 10 - 6 } { 15 } = \frac { 13 } { 15 }$.