Three coins are tossed. Find the probability that they all land heads up when at least one coin shows a head.
Number of elements in sample space $S = n \left(S\right) = 2^{3} = 8$
Let $A$ be the event that all 3 coins land heads up.
Then $A = \left\{H, \; H, \; H\right\}$
Let $B$ be the event that at least one coin shows a head.
Then $\overline{B}$ is the event that all coins land tails up.
Now, $\overline{B} = \left\{T, \; T, \; T\right\}$ $\implies$ $n \left(\overline{B}\right) = 1$
$\therefore \;$ $P \left(\overline{B}\right) = \dfrac{n \left(\overline{B}\right)}{n \left(S\right)} = \dfrac{1}{8}$
$\therefore \;$ $P \left(B\right) = 1 - P \left(\overline{B}\right) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$
$\left(A \cap B\right) = $ event that all coins show heads AND at least one coin shows a head
Now, $\left(A \cap B\right) = \left\{H, \; H, \; H\right\}$ $\implies$ $n \left(A \cap B\right) = 1$
$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{1}{8}$
$\therefore \;$ Probability that all coins land heads up when at least one coin shows a head
$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{1/8}{7/8} = \dfrac{1}{7}$