For

P (B) > 0

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} .

Hence

P (B_{j} ∣ A) = \frac{P ( A ∣ B _{j} ) P ( B _{j} )}{P ( A )} .

Because the

B_{k}

form a partition, the law of total probability gives

P (A) = \sum_{k = 1}^{n} P (A ∣ B_{k}) P (B_{k}) .

Substitution proves Bayes theorem in the required form:

P (B_{j} ∣ A) = \frac{P ( A ∣ B _{j} ) P ( B _{j} )}{\sum _{k = 1}^{n} P ( A ∣ B _{k} ) P ( B _{k} )} .

ii

Solution

Urn

r

has

n - r

blue balls among

n - 1

balls. Averaging over the uniformly chosen urn,

P (first blue) = \frac{1}{n} \sum_{r = 1}^{n} \frac{n - r}{n - 1} = \frac{1}{2} .

The probability that both removed balls are blue is

P (B_{1} \cap B_{2}) = \frac{1}{n} r = 1 \sum n \frac{( n - r ) ( n - r - 1 )}{( n - 1 ) ( n - 2 )} = \frac{1}{3} .

Therefore

P (B_{2} ∣ B_{1}) = \frac{P ( B _{1} \cap B _{2} )}{P ( B _{1} )} = \frac{2}{3} .

The first blue draw makes urns with more blue balls more likely, which explains why the conditional probability exceeds one half.

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