
Conditional Probability
Box A has 1 red and 2 white balls, whereas box B has 1 red and 1 white ball. Two balls are randomly selected, without replacement, from box A and placed in box B. If you now select a ball at random, from box B, what is the probability that is it red?
Im stuck as I dont know what method to use to go about this question.
Thanks

Well, there are two possibilities for our what our final box B looks like right before we draw from it.
First, we could have taken 1 red and 1 white from Box A. This occurs with probability 2/3.
Secondly, we could have taken 2 white balls from Box A. This occurs with probability 1/3.
Under the first scenario, we will be drawing from a box that has 2 red and 2 white, and so we will have a 2/4 or 1/2 chance to get a red ball.
Under the second scenario, we will be drawing from a box that has 1 red and 3 white balls. This gives us a 1/4 chance to get a red ball.
(2/3)(1/2)+(1/3)(1/4) = (1/3)+(1/12) = (5/12)