A conditional probability question: Three identical jewelry boxes with two draws each. Each draw contains a watch

Question

The following problem is from the book "Probability and Statistics" which is part of the Schaum's outline series. It can be found on page 30 and is problem number 1.57.
Problem:
Each of three identical jewlery boxes has two draws. In each draw of the first box there is a gold watch. In each each draw of the second box there is a silver watch. In one draw of the third box there is a gold watch while in the other there is a silver watch. If we select a box at random, open one of the drawers and find it contains a silver watch, what is the probability that the other drawer has the gold watch?
Answer:
Let $A_1$ be the probability that we selected the first box. Let $A_2$ be the probability that we selected the second box. Let $A_3$ be the probability that we selected the third box. Let $A$ be the probability that the first draw we open contains a silver watch. Now my goal will be to find the probabilities: $P(A_1)$, $P(A_2)$ and $P(A_3)$. Given these probabilities, I can find the probability that the other draw has a gold watch. Since the first box has only gold watches in this case, the first box was not selected. \begin{align*} P( A_2 | A ) &= \frac{ P(A_2)P( A|A_2) } { \sum_{\substack{j=1 }}^n { P(A_j)P(A|A_j) } } \\ P(A_1) &= P(A_2) = P(A_3) = \frac{1}{3} \\ P( A|A_1) &= 0 \\ P( A|A_2) &= 1 \\ P( A|A_3) &= \frac{1}{2} \end{align*} \begin{align*} P( A_2 | A ) &= \frac{ \left( \frac{1}{3} \right) (1) } { \sum_{\substack{j=1 }}^n { P(A_j)P(A|A_j) } } \\ 3 P( A_2 | A ) &= \frac{ 1 } { P(A_1)P(A|A_1) + P(A_2)P(A|A_2) + P(A_3)P(A|A_3) } \\ 3 P( A_2 | A ) &= \frac{ 1 } { \left( \frac{1}{3} \right) P(A|A_1) + \left( \frac{1}{3} \right) P(A|A_2) + \left( \frac{1}{3} \right) P(A|A_3) } \\ 3 P( A_2 | A ) &= \frac{ 3 } { P(A|A_1) + P(A|A_2) + P(A|A_3) } \\ P( A_2 | A ) &= \frac{ 1 } { 0 + 1 + \frac{1}{2} } \\ P( A_2 | A ) &= \frac{3}{2} \\ P( A_3 | A ) &= \frac{ P(A_3)P( A|A_3) } { \sum_{\substack{j=1 }}^n { P(A_j)P(A|A_j) } } \\ P( A_3 | A ) &= \frac{ P(A_3)P( A|A_3) } { P(A_1)P ( A | A_1 ) + P(A_2)P ( A | A_2 ) + P(A_3)P ( A | A_3 ) } \\ P( A_3 | A ) &= \frac{ \left( \frac{1}{3} \right) \left( \frac{1}{2} \right) } {\left( \frac{1}{3} \right)( 0 ) + \left( \frac{1}{3} \right)(1) + \left( \frac{1}{3} \right) \left( \frac{1}{2 }\right) } \\ P( A_3 | A ) &= \frac{ \frac{1}{6} } { \frac{3}{6} } \\ P( A_3 | A ) &= \frac{1}{3} \\ \end{align*} \begin{align*} P(A) &= P(A_1)P(A|A_1) + P(A_2)P(A|A_2) + P(A_3)P(A|A_3) \\ P(A) &= \left( \frac{1}{3 }\right) (0) + \frac{1}{3} \left( 1 \right) + \frac{1}{3} \left( \frac{1}{2}\right) \\ P(A) &= \frac{1}{3} + \frac{1}{6} \\ P(A) &= \frac{1}{2} \end{align*} The book's answer is $\frac{1}{3}$. Where did I go wrong? I am also thinking I did a lot of unnecessary work but I am not sure about that.

Answer 1

I am also thinking I did a lot of unnecessary work but I am not sure about that.

You have already chosen a silver drawer, so the gold drawers become irrelevant.

The issue is now, of the relevant three drawers, what is the probability that your random choice is the one drawer that was paired with gold?

The answer is obviously 1/3.