You have three cards in a hat. One is white on both sides, one is red on both sides, the third is white on one side and red on the other. You pull a card out of the hat and lay it down. The face is white. What are the chances that the other side of the card is also white?
Edit: to those who have marked this as a duplicate, I invite you to explain how it can be so when the answer in the "duplicate" is different from this one.
Answer
I think the correct answer is
2/3
We use the law of conditional probability.
Let $WW$ be the event that we get the all-white card, $RW$ the mixed card and $RR$ the all-red card. Let $W$ be the event that we get a white face up when we pull a card from the bag.
The law of conditional probability gives us $$P(WW \mid W) = \frac{P(WW \cap W)}{P(W)} = \frac{1/3}{1/2} = \frac{2}{3}$$ as $P(WW \cap W)$ simply reduces to the probability that we drew the all-white card, and half the faces are white, so $P(W)$ is a half.
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