Answer to Question #235407 in Statistics and Probability for Olu

Question

A prisoner is considering his chances of escape .From his cell, there are 3 possible exits 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 , what is the probability that the prisoner will make a successful escape (to 3 decimal places)

i.0.333
ii.0.667
iii.0.265
iv.0.633
v.0.367

Answer 1

P(exit A) = P(exit B) = P(exit C) $= \frac{1}{3}$

P(Successful escape | exit A) = 0.4

P(Successful escape | exit B) = 0.3

P(Successful escape | exit C) = 0.4

There are three possible exits.

Hence, the probability of choosing any one exit will be $\frac{1}{3}$

Now, the probability will make a successful escape if he either chooses exit A and escape successfully or chooses exit B and escape successfully or chooses exit A and escape successfully.

P(Escaped successfully) $= \frac{1}{3} \times (0.4 + 0.3 + 0.4)$

P(Escaped successfully) $= \frac{1.1}{3}$

P(Escaped successfully) = 0.367

Hence, the probability that the prisoner will make a successful escape is 0.367.

Option (V) is correct.