# Statistics question Conditional Probability

Question: Of three cards, one is painted red on both sides; one is painted black on both sides; and one is painted red on one side and black on the other. A card is randomly chosen and placed on a table. If the side facing up is red, what is the probability that the other side is also red?

My Attempt: number designates side and letter the colour Card 1: R1 R2 Card 2: B1 B2 Card 3: R1 B2 P(R2|R1) = P(R1R2)/P(R1) P(R2|R1) = (1/3)/(1/2) P(R2|R1) = 2/3 I did this in a conditional probability manner but my instinct says the answer should just be 1/2...

• Your intuition is wrong, and your reasoning is correct! Commented Oct 11, 2014 at 21:33

Your intuition is that each card has an equal probability of being chosen, and this is true.   Yet you must consider that the cards have an not so equal probability of being chosen and showing their red side.

Instead, observe that if I select a card, then select a side to show, both choices without bias, then every side has an equal probability of being the one shown.

Now, when given that the side shown is red, the three red sides still have equal probability of being that one shown.   However only of them have a red otherside.   The third red side has a green otherside.

Therefore there must be a conditional probability of $2/3$ for the otherside of the side shown to be red when given that the side shown is itself red.

Listing the possibilities is easy here and gives the right intuition. Number the sides of the double-red card. If the top is red, there are three possibilities (not 2):

1. The red1/red2 card was chosen with 1 on top.
2. The red1/red2 card was chosen with 2 on top.
3. the red/black card was chosen with red on top.

This gives a probability of 2/3 despite there being only 2 cards that could have been chosen, because you condition on choosing among the red sides, not the cards.

• thanks, this was very clear. Commented Oct 11, 2014 at 21:43