Question

Odds $8$ to $5$ against a person who is $40$yr old living till he is $70$ and $4$ to $3$ against another person now $50$ till he will be living $80$. Probability that one of them will be alive next $30$yr.

• A. $\frac{59}{91}$
• B. $\frac{44}{91}$
• C. $\frac{51}{91}$
• D. $\frac{32}{91}$

Answer 1

The correct option is B $\frac{44}{91}$

let A be the event that person A will die in 30 yrs and let B be the event that person B will die in 30 yrs.

$P\left(A\right)=\frac{8}{8+5}$

$P\left(A\right)=\frac{8}{13}\Rightarrow P\left(\overline{A}\right)=\frac{5}{13}$

Similarly, $P\left(B\right)=\frac{4}{7}$ or $P\left(\overline{B}\right)=\frac{3}{7}$

There are two ways in which one person is alive after $30$ yr.

$\overline{A}B$ and $A\overline{B}$ and event are independent.

So, required probability
$=P\left(\overline{A}\right)\cdot P\left(B\right)+P\left(A\right)\cdot P\left(\overline{B}\right)$

$=\frac{5}{13}\cdot \frac{4}{7}+\frac{8}{13}\cdot \frac{3}{7}=\frac{44}{91}$.