Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

closed as off-topic by Amzoti, azimut, Donkey_2009, TMM, Nicholas R. Peterson Oct 15 '13 at 19:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Amzoti, azimut, Donkey_2009, TMM, Nicholas R. Peterson
If this question can be reworded to fit the rules in the help center, please edit the question.

4  
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

2 Answers 2

up vote 0 down vote accepted

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

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

share|improve this answer
    
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.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.