91Ó°ÊÓ

The variance of the length of time until the miner reaches safety.

Let's calculate second moment using the similar idea in Example $5c$We have that,

$E\left(X^2\right)=E\left(X^2âˆ\pounds Y=1\right)P(Y=1)+E\left(X^2âˆ\pounds Y=2\right)P(Y=2)$

$+E\left(X^2âˆ\pounds Y=3\right)P(Y=3)$

Observe that,

$E\left(X^2âˆ\pounds Y=1\right)=9$.

Since if he chooses the first door, the expected square of time needed to escape is equal to $9$. Also, if he chooses the second door, he will need $5+X$of time to escape. Similarly if he chooses the third door.

Hence, $E\left(X^2âˆ\pounds Y=2\right)=E\left((X+5{)}^2\right)$$=E\left(X^2\right)+10E\left(X\right)+25$

$E\left(X^2âˆ\pounds Y=3\right)=E\left((X+7{)}^2\right)$$=E\left(X^2\right)+14E\left(X\right)+49$.

Therefore, we end up with equation

$E\left(X^2\right)=\frac{1}{3}\left(9+E\left(X^2\right)+10E\left(X\right)+25+E\left(X^2\right)+14E\left(X\right)+49\right)$

Which yields,

$E\left(X^2\right)=83+24E\left(X\right)$

Add the value,

$=443$

Because we know that $E\left(X\right)=15$.

The variance is equal to$\left(X\right)=E\left(X^2\right)-E{\left(X\right)}^2$

$=218$.

The required variance is equal to value are$218$.