2.1.2 problem 4

(a)

$$\begin{array}{cccc}P\left(K\right|R)& =\frac{P\left(K\right)P\left(R\right|K)}{P\left(R\right)}& & \\ & =\frac{P\left(K\right)P\left(R\right|K)}{P\left(K\right)P\left(R\right|K)+P(K^c\left)P\right(R\left|K^c\right)}& & \\ & =\frac{p}{p+(1-p)\frac{1}{n}}& & \end{array}$$

(b)

Since $p+(1-p)\frac{1}{n}\le 1$, $P\left(K\right|R)\ge p$ with strict equality only when $p=1$. This result makes sense, since if Fred gets the answer right, it is more likely that he knew the answer.