# Card probability problem [duplicate]

Possible Duplicate:
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?

• This has to be a duplicate. May 16, 2012 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, 2012 at 19:05
• @David here it is again
– MJD
May 16, 2012 at 19:07

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 $P(DB\cap OB)=P(DB)=\frac{1}{3}$.
• 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? May 16, 2012 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. May 16, 2012 at 19:18