Question

Supposing that it is $9$ to $7$ against a person $A$ who is now $35$ years of age living till he is $65$, and $3$ to $2$ against a person $B$ now $45$ till he is $75$; find the chance that one at least of these persons will be alive $30$ years hence.

• A. $\frac{27}{80}$
• B. $\frac{14}{80}$
• C. $\frac{66}{80}$
• D. $\frac{53}{80}$

Answer 1

Answer: (D)

$\frac{53}{80}$

The chance that A will die within 30 years $=\frac{9}{16}$
The chance that B will die within 30 years =$=\frac{3}{5}$
$\therefore$ the chance that both will die is $\frac{9}{16}\times \frac{3}{5}=\frac{27}{80}$
$\therefore$ the chance that both will not be dead, that is, one at least will be alive is $1-\frac{27}{80}=\frac{53}{80}$