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Exercise 3.31

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(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 (X_{i}) = \frac{(\binom{3}{i}) (\binom{3}{3 - i})}{(\binom{6}{3})}

Thus, $P (X_{2}) + P (X_{3}) = \frac{10}{(\binom{6}{3})} = \frac{1}{2}$

(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 (M | T)}{P (M^{c} | T)} = \frac{P (M)}{P (M^{c})} \frac{p_{1}}{1 - p_{2}} = \frac{p_{1}}{1 - p_{2}}

fifthist

2021-12-05 00:00

Exercise 3.31

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