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Three people each flip two fair coins.Find the probability that exactly two of the people flipped one head and one tail.


Out of three persons,two persons can be chosen in $\binom{3}{2}$ ways.Each person flips two fair coins.So each persons gets $HH,HT,TH,TT$.Probability of a person getting one head and one tail is $\frac{1}{2}$.

So the probability that exactly two of the people flipped one head and one tail is $\binom{3}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{3}{4}$

But my answer is wrong.What is wrong in my approach.Please help me.

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    $\begingroup$ Multiply by $1/2$ for the third person to flip double head or double tail. $\endgroup$ – André Nicolas Dec 1 '15 at 3:45
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As you say, the probability of a person getting one head and one tail is $\frac12$. The probability of getting two heads or two tails is $\frac12$. Altogether, we'll have:

$${3 \choose 2}\times\left(\frac12\right)^2\times\frac12 = \frac38.$$

The $\left(\frac12\right)^2$ accounts for the two people who get a head and a tail, and the final $\frac12$ accounts for the one person who doesn't.

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You accounted for the first person flipping a head and a tail, the second person flipping a head and a tail, but you didn't account for the third person not flipping a head and a tail.

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