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Find the probability that, in three tosses of a fair coin, there are three heads, given that there is at least one head.

I manage to get $\frac{3}{6}$ or $\frac{1}{6}$ but the right answer is


I have no idea, Can you please explain?

thanks! Appraciate it!

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How many outcomes are there with at least one head? How many of those have three heads? – Gerry Myerson Aug 4 '12 at 10:40
@GerryMyerson 1/6 and 3/6=1/2 ? it is what u mean? thx – Sb Sangpi Aug 4 '12 at 10:53
6 outcomes? Can you list them? – Gerry Myerson Aug 4 '12 at 12:25
up vote 4 down vote accepted

A fair coin has etqual probabilitiers for heads and tails. Therefore the probability can simply be found by counting.

There are eight different possibilities for outcomes of three tosses:

  1. Head, Head, Head
  2. Head, Head, Tail
  3. Head, Tail, Head
  4. Head, Tail, Tail
  5. Tail, Head, Head
  6. Tail, Head, Tail
  7. Tail, Tail, Head
  8. Tail, Tail, Tail

Possibilities 1 to 7 have at least 1 head, while possibility 8 doesn't. Therefore remove that last one.

From the 7 remaining possibilities, 1 has all three heads, so the probability is one out of seven, or $\frac17$.

OK, but what to do if there are too many cases to explicitly list them? Well, it is easy to see that you cannot have three heads without having at least one head. So your probability is $$p_{\text{3 heads if at least 1 head}} = \frac{\text{Number of cases with 3 heads}}{\text{Number of cases with at least 1 head}}$$ However you know that $$p_{\text{3 heads}} = \frac{\text{Number of cases with 3 heads}}{\text{Number of all cases}}$$ and $$p_{\text{at least 1 head}} = \frac{\text{Number of cases with at least 1 head}}{\text{Number of all cases}}$$ From this, it is not hard to see that $$p_{\text{3 heads if at least 1 head}} =\frac{p_{\text{3 heads}}}{p_{\text{at least 1 head}}}$$ Now you surely know that $p_{\text{3 heads}}=\left(\frac12\right)^3=\frac18$ and $p_{\text{at least 1 head}} = 1-p_{\text{3 tails}}=\frac78$. Inserting then again gives $p_{\text{3 heads if at least 1 head}}=\frac18/\frac78=\frac17$.

Indeed, that latter formula works even for arbitrary probabilities (i.e. not only fair coins). Note however that it still depends on the fact that having 3 heads implies having at least 1 head.

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Probability of 3 heads = ${\left(\frac{1}{2}\right)}^3 =p_1$(say)

Probability of at least 1 heads = 1 - probability of 0 heads = $1 - \left(1-\frac{1}{2}\right)^3 =\frac{7}{8}=p_2$ (say)

Applying conditional probability, the desired probability is $\frac{P(A∩B)}{P(B)} = \frac{p_1}{p_2}$ as here $P(A∩B) = P(A)$,


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