Answer to Question #235712 in Statistics and Probability for Tyza

Question

A prisoner is considering his chances of escape. From his cell, there are 3 possible exists he could choose (exit A, exit B, exit C). From these exits, the chances of successfully escaping are 0.4, 0.3 and 0.4 respectively.

If the prisoner picks an exit at random, what is the probability that the prisoner will make a successful escape (to 3 decimal places)?

Answer 1

we have the following information;

P(exit A)= 1/3

P(exit B)=1/3

P(exit C)=1/3

Let S be the event that the prisoner successfully escapes, then

P(S| exit A)=0.4

P(S| exit B)=0.3

P(S| exit C)=0.4

We want to determine the probability that the prisoner escapes successfully, P(S).

We use the law of total probability to find P(S) as illustrated below.

P(S)=P(S| exit A)*P(exit A)+P(S| exit B)*P(exit B)+P(S| exit C)*P(exit C)

=0.4*(1/3)+ 0.3*(1/3)+ 0.4*(1/3)

=0.367