Find the probability of three tosses of a fair coin 
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

$\frac{1}{7}$ 

I have no idea, Can you please explain?
thanks! Appraciate it!
 A: 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:


*

*Head, Head, Head

*Head, Head, Tail

*Head, Tail, Head

*Head, Tail, Tail

*Tail, Head, Head

*Tail, Head, Tail

*Tail, Tail, Head

*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.
A: 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 $P(A∩B) = P(A)$,
$$\frac{p_1}{p_2}=\frac{\frac{1}{8}}{\frac{7}{8}}=\frac{1}{7}$$
A: Let $A$ be the event that all events succeed and let $B$ denote the event that at least one event succeeds.
Since the event $B$ is a subset of $A$ it follows that
$P(A \mid B) = \frac{P(A, B)}{P(B)} = \frac{P(A)}{P(B)}$.
Further, $P(B^*) = P(A)$ so the answer is $\frac{p^n}{1-p^n}$.
The probability of success is $p=1/2$ and since you toss $n=3$ coins the answer is $1/7$.
Additional note: In the special case when $p = 1 / k$ we get $1 / (k^n - 1)$ which again gives $1/(2^3 - 1)$.
