Answer to Question #233648 in Statistics and Probability for Uyu

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.The prisoner picks an exit at random. If he did make a successful escape, what is the probability that the escape was made via exit A (final calculation to 3 decimal places)?

Answer 1

We have following informations :

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

Now, probability will make a success 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.

We have to find P(exit A | successful escape).

P(exitA|successful escape) $= \frac{P(exit \;A\; and \;successful \;escape)}{P(successful \;escape)}$

P(exitA|successful escape) $= \frac{P(sucessful \;escape | exit \;A) \times P(exit \;A)}{P(successful \;escape)}$

$= \frac{0.4 \times (1/3)}{0.367} = 0.364$