# Bayes Rule Example

Suppose that we have two bags each containing black and white balls. One bag contains three times as many white balls as blacks. The other bag contains three times as many black balls as white. Suppose we choose one of these bags at random. For this bag we select five balls at random, replacing each ball after it has been selected. The result is that we find 4 white balls and one black. What is the probability that we were using the bag with mainly white balls?

Solution. Let A be the random variable "bag chosen" then A={a1,a2} where a1 represents "bag with mostly white balls" and a2 represents "bag with mostly black balls" . We know that P(a1)=P(a2)=1/2 since we choose the bag at random.

Let B be the event "4 white balls and one black ball chosen from 5 selections".

Then we have to calculate P(a1|B). From Bayes' rule this is: Now, for the bag with mostly white balls the probability of a ball being white is ¾ and the probability of a ball being black is ¼. Thus, we can use the Binomial Theorem, to compute P(B|a1) as: Similarly hence 