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Three cards in an urn bear pictures of ants and bees; one card has ants on both sides, and one card has bees on bath sides, and one has an ant on one side and a bee on the other. A card is removed at random and places flat. If the upper face shows a bee, what is the probability that the other side shows an ant?

If we let $A$ be the event that the lower side shows an ant and $B$ be the event that the upper face shows a bee.

My first port of call is:

$$P(A |B) = \frac{P(A \cap B)}{P(B)} = \frac{1/3}{P(B)}$$

But I don't think $P(B)$ is $2/3$. I can't explain why. The answer according to the book is $1/3$

HMPtwo
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  • @ParclyTaxel the link provided only works if the front and the back sides of a card are not distinguishable from each other. Don't think it's true about playing cards, and here not clear what OP actually means... – gt6989b Oct 03 '22 at 15:52
  • @gt6989b playing cards don't have pictures of ants and bees on them... these are not standard playing cards. Given what all has been described so far, I see no reason to think that this is any different of a problem. – JMoravitz Oct 03 '22 at 16:01
  • Of the three bees, two of them are on the double bee card and only one is on the ant-bee card. Therefore, there is a two thirds chance that there is no ant on the other side which means there is only a one third chance that there is. – John Douma Oct 03 '22 at 16:26
  • Following the same logic. If we let $X_{AA}$ be the card with ant on both sides, $X_{BA}, X_{BA}$ similarly defined. Then $P(B) = P(B|X_{AA})P( X_{AA}) + P(B|X_{AB}) P(X_{AB}) +P(B|X_{BB}) P(X_{BB}) = 1/6 + 1/3 $. so $P(A|B ) = 2/3$ – HMPtwo Oct 03 '22 at 20:50

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