Question

A prisoner is considering his chances of escape. From his cell, there are 3 possible exits he could choose and if he chooses a particular exit, the probability he will escape is 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

$$P(1)P(Yes|1)+P(2)P(Yes|2)+P(3)P(Yes|3)$$

$$=\dfrac{1}{3}(0.4)+\dfrac{1}{3}(0.3)+\dfrac{1}{3}(0.4)=\dfrac{11}{30}\approx0.367$$