# Probability Question Involving Coin Tossing [closed]

Two coins are tossed. What is the conditional probability that two heads result given that there is at least one head?

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## closed as off-topic by Amzoti, azimut, Donkey_2009, TMM, Nick PetersonOct 15 '13 at 19:53

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Leon - I think you need to start showing effort by posting your thoughts and attempts at solving your homework questions (as has been repeatedly recommended to you in earlier comments). This site is not meant to be a "do your homework for you" site. By now, you should know that: You've asked $13$ questions now, on most of which failed to show any work on your part, and for none of which have you accepted any answers. –  amWhy Oct 15 '13 at 18:32
I never knew there were any particular rules to Stack Exchange. I'm sorry if I've offended you. –  Vladimir Nabokov Oct 15 '13 at 18:37

$P = \dfrac{(1/2)(1/2)}{1 - 1/4} = \dfrac{1}{3}$

Let $E$ be the event that both of them are heads and $F$ be the event that at least one of them is heads. Then it follows that $P(F)=\frac{3}{4},P(E\cap F)=P(E)=\frac{1}{4}.$ Then $P(E|F)=\frac{1}{3}.$

Recall that $P(E\cap F)=P(F)P(E|F).$

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How did you get to that result? –  Vladimir Nabokov Oct 15 '13 at 18:36

$P\left\{ X=2\mid X\geq1\right\} =\frac{P\left\{ X=2\wedge X\geq1\right\} }{P\left\{ X\geq1\right\} }=\frac{P\left\{ X=2\right\} }{1-P\left\{ X=0\right\} }=\frac{\frac{1}{2}\times\frac{1}{2}}{1-\frac{1}{2}\times\frac{1}{2}}=\frac{1}{3}$

Here $X$ stands for the number of heads.

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