Answer to Question #235835 in Statistics and Probability for Ammarah

Question

Questions 1 and 2 refer to the following scenario. 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)? A. 0.333 B. 0.367 C. 0.265 D. 0.633 E. 0.667 Question 2. 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)? A. 0.364 B. 0.429 C. 0.133 D. 0.333 E. 0.571

Answer 1

1. By the Law of Total Probability

$$P(S)=P(S|A)(P(A)+P(S|B)P(B)+P(S|C)P(C)$$

$$=0.4(\dfrac{1}{3})+0.3(\dfrac{1}{3})+0.4(\dfrac{1}{3})=\dfrac{1.1}{3}\approx0.367$$

$B. 0.367$

2. By Bayes’ Theorem

$$P(A|S)=\dfrac{P(S|A)(P(A)}{P(S|A)(P(A)+P(S|B)P(B)+P(S|C)P(C)}$$

$$=\dfrac{0.4(\dfrac{1}{3})}{\dfrac{1.1}{3}}=\dfrac{4}{11}\approx0.364$$

$A. 0.364$