In Example 5c, compute the variance of the length of time until the miner reaches safety.

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,

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

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

Observe that,

$E\left(X^2\mid Y=1\right)=9$.

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^2\mid Y=2\right)=E\left((X+5{)}^2\right)$$=E\left(X^2\right)+10E\left(X\right)+25$

$E\left(X^2\mid 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$218$.