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 $45$ living till he is $75;$ find the chance that one at least of these persons will be alive $30$ years hence.

Answer 1

$\frac{P\left(A^{{}^{\prime }}\right)}{P\left(A\right)}=\frac{9}{7}$ for $65$ years $P\left(A\right)=\frac{7}{16}$
$\frac{P\left(B^{{}^{\prime }}\right)}{P\left(B\right)}=\frac{3}{2}$ For $75$ years $P\left(B\right)=\frac{2}{5}$
$P\left(B^{{}^{\prime }}\right)=\frac{3}{5}$
Required probability$=P\left(A\right)P\left(B^{{}^{\prime }}\right)+P\left(A^{{}^{\prime }}\right)P\left(B\right)+P\left(A\right)P\left(B\right)$
$=\frac{7}{16}\times \frac{3}{5}+\frac{9}{16}\times \frac{3}{5}+\frac{7}{16}\times \frac{2}{5}$
$\Rightarrow \frac{21}{80}+\frac{9}{40}+\frac{7}{40}$
$\Rightarrow \frac{21+18+14}{80}\Rightarrow \frac{53}{80}$