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A prisoner is trapped in a cell containing3doors. The first door leads to a tunnel that returns him to his cell after 2days鈥 travel. The second leads to a tunnel that returns him to his cell after 4 days鈥 travel. The third door leads to freedom after 1day 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?

Short Answer

Expert verified

The expected number of days until the prisoner reaches freedom is

E[X]=12.

Step by step solution

01

Given information

Given in the question that, a prisoner is trapped in a cell containing3doors. The first door leads to a tunnel that returns him to his cell after 2days鈥 travel. The second leads to a tunnel that returns him to his cell after 4days鈥 travel. The third door leads to freedom after 1day of travel. If it is assumed that the prisoner will always select doors1,2,and 3with respective probabilities .5,.3,and .2,what is the expected number of days until the prisoner reaches freedom?

02

Explanation

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

localid="1647510264291" E(X)=.5E[XY=1]+.3E[XY=2]+.2E[XY=3]

Now, E[XY=1]=E[X]+2, since the prisoner will essentially get back to the cell and the issue begins once again.

Similarly, E[XY=2]=E[X]+4.E[XY=3]=1, of course. Hence,

E[X]=.5(E[X]+2)+.3(E[X]+4)+.2=.5E[X]+1+.3E[X]+1.2+.2=.8E[X]+2.4

E[X]=12.

03

Final answer

The expected number of days until the prisoner reaches freedom is

E[X]=12.

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