We need to determine P(coin1|head). By applying Bayes' theorem,

P(coin1 | head) = $\frac{P(\text{coin1}) \times P(\text{head}|\text{coin1})}{(P(\text{coin1}) \times P(\text{head}|\text{coin1}) + P(\text{coin2}) \times P(\text{head}|\text{coin2}))}$ = $\frac{(1 \times \frac{1}{2})}{(\frac{1}{2} \times 1 + \frac{1}{2} \times \frac{1}{2})}$ = $\frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{4}}$ = $\frac{1/2}{3/4}$ = $\frac{2}{3}$

Key points:

• P(coin1) = P(coin2) = $\frac{1}{2}$ since we have two coins which are equally likely
• P(head|coin1) = 1 since coin 1 is double headed
• P(head|coin2) = $\frac{1}{2}$ since coin 2 is unbiased (both head and tail are equally likely)