Q.53 - A First Course in Probability

Question

A prisoner is trapped in a cell containing$3$doors. The first door leads to a tunnel that returns him to his cell after $2$days’ travel. The second leads to a tunnel that returns him to his cell after $4$ days’ travel. The third door leads to freedom after $1$day of travel. If it is assumed that the prisoner will always select doors $1, 2,$ and $3$ with respective probabilities $.5, .3,$ and .$2$, what is the expected number of days until the prisoner reaches freedom? 

Step-by-Step Solution

Answer

The expected number of days until the prisoner reaches freedom is 

 $\mathbb{E}\left[X\right]=12$.

Step 1: Given information

Given in the question that, a prisoner is trapped in a cell containing$3$ doors. The first door leads to a tunnel that returns him to his cell after $2$ days’ travel. The second leads to a tunnel that returns him to his cell after $4$ days’ travel. The third door leads to freedom after $1$day of travel. If it is assumed that the prisoner will always select doors$1, 2,$ and $3$ with respective probabilities $.5, .3,$ and .$2,$what is the expected number of days until the prisoner reaches freedom?

Step 2: Explanation

Allow $X$ to signify the quantity of days until the detainee get away and $Y$ mean the entryway the detainee picks. Then

$E\left(X\right)=.5·E[X\mid Y=1]+.3·E[X\mid Y=2]+.2·E[X\mid Y=3]$

Now, $E[X\mid Y=1]=E\left[X\right]+2$, since the prisoner will essentially get back to the cell and the issue begins once again.

Similarly, $E[X\mid Y=2]=E\left[X\right]+4.E[X\mid Y=3]=1$, of course. Hence,

$E\left[X\right]=.5\left(E\right[X]+2)+.3\left(E\right[X]+4)+.2=.5·E\left[X\right]+1+.3·E\left[X\right]+1.2+.2=.8·E\left[X\right]+2.4$

$\mathbb{E}\left[X\right]=12$. 

Step 3: Final answer

The expected number of days until the prisoner reaches freedom is 

 $\mathbb{E}\left[X\right]=12$.