Answer to Question #103253 in Statistics and Probability for Christopher Cuya

Question

Recall the puzzle about the prisoner and the king. Imagine that the prisoner puts 1 white ball in one box, and 14 white and 15 black balls in the other box. Then the king chooses a random box and then chooses a random ball inside this box. What is the conditional probability of the event "the king chooses the second box" under the condition "the ball chosen by the king was white"?

Answer 1

Let A denote the event that a first box is selected.

Let B denote the event that a second box is selected.

$$P(A)={1 \over 2}, P(B)={1 \over 2}$$

Then

$$P(White|A)=1, P(White|B)={14 \over 14+15}={14 \over 29}$$

Bayes' rule shows

$$P(B|White)={P(White|B)P(B) \over P(White|A)P(A)+P(White|B)P(B)}$$

$$P(B|White)={{14 \over 29} ({1 \over 2} ) \over 1({1 \over 2})+{14 \over 29} ({1 \over 2})}$$

$$P(B|White)={14 \over 43}\approx0.3256$$