Question

Odds $8$ to $5$against a person who is $40$ years 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 then will be alive next $30$ years

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

Answer 1

The correct option is B

$\frac{44}{91}$

Explanation for the correct option(s)

Find the probability.

Let $A$be the event that $A$ will be die in $30$ years

Let $B$ be the event that $B$ will be die in $30$ years

The probability of $A$can be expressed as,

$$\begin{gathered} P\left(A\right)=\frac{8}{8+5}=\frac{8}{13} \\ P\left(\overline{A}\right)=\frac{5}{13} \end{gathered}$$

The probability of $B$can be expressed as,

$$\begin{gathered} P\left(B\right)=\frac{4}{4+3}=\frac{4}{7} \\ P\left(\overline{B}\right)=\frac{3}{7} \end{gathered}$$

We know that,

$$\begin{gathered} P\left(A\cap \overline{B}\right)=P\left(A\right).P\left(\overline{B}\right) \\ P\left(\overline{A}\cap B\right)=P\left(\overline{A}\right).P\left(B\right) \end{gathered}$$

Then,

$$\begin{gathered} P\left(A\cap \overline{B}\right)\cap P\left(\overline{A}\cap B\right)=P\left(A\right).P\left(\overline{B}\right)+P\left(\overline{A}\right).P\left(B\right) \\ =\frac{8}{13}\times \frac{3}{7}+\frac{5}{13}\times \frac{4}{7} \\ =\frac{24}{91}+\frac{20}{91} \\ =\frac{44}{91} \end{gathered}$$

Hence, option B is correct .