7. [4] George has two coins, one of which is fair and the other of which always comes up heads. Jacob takes one of them at random and flips it twice. Given that it came up heads both times, what is the probability that it is the coin that always comes up heads?
Answer: $\sqrt[4]{5}$ In general, $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$, where $P(A \mid B)$ is the probability of $A$ given $B$ and $P(A \cap B)$ is the probability of $A$ and $B$ (See http://en.wikipedia.org/wiki/Conditional_probability for more information). If $A$ is the event of selecting the "double-headed" coin and $B$ is the event of Jacob flipping two heads, then $P(A \cap B)=\left(\frac{1}{2}\right)(1)$, since there is a $\frac{1}{2}$ chance of picking the doubleheaded coin and Jacob will always flip two heads when he has it. By conditional probability, $P(B)=$ $\left(\frac{1}{2}\right)(1)+\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)=\frac{5}{8}$, so $P(A \mid B)=\frac{1 / 2}{5 / 8}=\frac{4}{5}$.