3.2.12 problem 31

(a)

Note that the distribution is not Binomial, since the guesses are not independent of each other. If, for instance, the woman guesses the first three cups to be milk-first, and she is correct, then the probability of her guessing milk-first on subsequent guesses is $0$, since it is known in advance that there are only $3$ milk-first cups.

Hypergeometric story fits. Let $X_i$ be the probability that the lady guesses exactly $i$ milk-first cups correctly.

$$P\left(X_i\right)=\frac{\left(\frac{3}{i}\right)\left(\frac{3}{3-i}\right)}{\left(\frac{6}{3}\right)}$$

Thus, $P\left(X_2\right)+P\left(X_3\right)=\frac{10}{\left(\frac{6}{3}\right)}=\frac{1}{2}$

(b)

Let $M$ be the event that the cup is milk first, and let $T$ be the event that the lady claims the cup is milk first. Then,

$$\frac{P\left(M\right|T)}{P\left(M^c\right|T)}=\frac{P\left(M\right)}{P\left(M^c\right)}\frac{p_1}{1-p_2}=\frac{p_1}{1-p_2}$$