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Probability problem

I found the following problem in Rosen's Discrete Mathematics and Its Applications 6th ed.:

There are three cards in a box. Both sides of one card are black, both sides of one card are red, and the third card has one black side and one red side. We pick a card at random and observe only one side.

If the side is black, what is the probability that the other side is also black?

Shouldn't the answer to the question be $1/3$, as among $3$ cards only $1$ of them has both sides black. But the book says answer is $2/3$, do I miss anything?

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marked as duplicate by MJD, Phira, Leonid Kovalev, Asaf Karagila, t.b. Aug 16 '12 at 12:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

This has to be a duplicate. –  David Mitra May 16 '12 at 19:01
You might also try performing the experiment, and seeing what happens. Or you could write a computer program to generate some random numbers and perform the experiment for you. –  MJD May 16 '12 at 19:05
@David here it is again –  MJD May 16 '12 at 19:07
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1 Answer

up vote 3 down vote accepted

An informal way of seeing it is that there are $6$ sides, and we are observing a randomly chosen side. Of the $3$ sides that are black, $2$ belong to the double-black card.

We can alternately do a formal conditional probability calculation. Let $DB$ be the event that the card picked is a double-black, and let $OB$ be the event that we observe a black. We want $P(DB|OB)$. By the usual formula, $$P(DB|OB)P(OB)=P(DB\cap OB).$$ We have $P(OB)=\frac{1}{2}$ and $P(DB\cap OB)=P(DB)=\frac{1}{3}$.

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Let $A$ be the event the second card is black and $B$ be the event first card is black. So $p(A|B) = p(A \cap B)/p(B)$. Here $p(A \cap B) = 1/3$ and $p(B) = 1/2$. Therefore $p(A|B) = 2/3$. Previously I mistakenly computed $p(A \cap B)$ instead. All correct now? –  f.nasim May 16 '12 at 19:12
@sven.hedin: Yes, perfectly correct. By the way, intuition should show that the answer of $1/3$ could not be right, since observing a black side increases the chances that we are dealing with the double-black. –  André Nicolas May 16 '12 at 19:18
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