Question

If the probability that A will live 15 years is $(\frac{7}{8})$ and that B will live 15 years is $(\frac{9}{10})$, then what is the probability that both will live after 15 years?

• A. $\frac{1}{20}$
• B. $\frac{63}{80}$
• C. $\frac{1}{5}$
• D. None of these

Answer 1

Answer: (B)

$\frac{63}{80}$

Answer 2

The correct option is (B): $\frac{63}{80}$
Explanation: To find the probability that both A and B will live for more than 15 years, we first need to determine the probabilities that each will not live for that duration.
1. The probability that A will not live for 15 years is:
$P(A \text{ not living}) = 1 - P(A \text{ living}) = 1 - \frac{7}{8} = \frac{1}{8}$
2. The probability that B will not live for 15 years is:
$P(B \text{ not living}) = 1 - P(B \text{ living}) = 1 - \frac{9}{10} = \frac{1}{10}$
Next, we can find the probability that at least one of them does not live for 15 years. This can be calculated using the formula for the probability of the union of two events:
$P(A \text{ or } B \text{ not living}) = P(A \text{ not living}) + P(B \text{ not living}) - P(A \text{ not living}) \cdot P(B \text{ not living})$
Substituting in the values:
$P(A \text{ or } B \text{ not living}) = \frac{1}{8} + \frac{1}{10} - \left(\frac{1}{8} \cdot \frac{1}{10}\right)$
Calculating each term:
$\frac{1}{8} + \frac{1}{10} = \frac{5}{40} + \frac{4}{40} = \frac{9}{40}$
$P(A \text{ not living}) \cdot P(B \text{ not living}) = \frac{1}{80}$
Thus:
$P(A \text{ or } B \text{ not living}) = \frac{9}{40} - \frac{1}{80}$
To subtract, convert $\frac{9}{40}$ to a fraction with a denominator of 80:
$\frac{9}{40} = \frac{18}{80}$
Now:
$P(A \text{ or } B \text{ not living}) = \frac{18}{80} - \frac{1}{80} = \frac{17}{80}$
Finally, the probability that both A and B will live after 15 years is:
$P(A \text{ and } B \text{ living}) = 1 - P(A \text{ or } B \text{ not living}) = 1 - \frac{17}{80} = \frac{63}{80}$
Therefore, the answer is B: $\frac{63}{80}$.