\"There are three identical cards that differ only in color. Both sides of one a

Question

"There are three identical cards that differ only in color. Both sides of one are black, both sides of the second one are red, and one side of the third card is black and its other side is red. These cards are mixed up and one of them is selected at random. If the upper side of this card is red, what is the probability that its other side is black?"

My answer is below. It disagrees with the Chegg (http://www.chegg.com/homework-help/fundamentals-of-probability-with-stochastic-processes-3rd-edition-chapter-3.4-problem-10e-solution-9780131453401) and solution manual answers. Please explain why my logic is wrong.

My thoughts are that there are two cards with red: the red/red and the red/black. Since the front of the card is red, we know that it must be one of those two cards--the sample space is those two cards. Only one of these is black. Therefore, only that one fulfils our desired event. So there's one possibility out of those two--and the probability is 1/2.

However, Chegg says it's 1/3. Please explain why my logic is wrong.

Explanation / Answer

This is wrong If you pick a card, and it has red on one side, 2/3s of the time it's going to be the card with two red sides, and 1/3 of the time it's going to be the card with the black side.

So the probability of seeing black on the other side given that you've seen red on one side is going to be 1/3. Considering the symmetry of B and R, calculating P(B|R) from P(R|B) is a bad plan to start with since you know they're the same.

Instead let X be the event you got the card with one color on each side, and Y the event that you see red on the side you're looking at.

P(X|Y) = P(Y and X) / P(Y)

P(Y and X) = 1/6 by considering there are six sides you could see, and exactly one of them is the red side on the multicolored card. P(Y) = 1/2

P(X|Y) = ( 1 / 6 ) / ( 1 / 2 )

= 2 / 6

= 1 / 3

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