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Here is the problem:

Each of 2 cabinets identical in appearance has 2 drawers. One of the cabinets has a silver coin in each drawer. The other cabinet has a silver coin in one drawer and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is revealed. What is the probability that the other drawer of the cabinet holds a silver coin?

And here's what I did:

Let $C1$ = event that cabinet 1 is chosen

Let $C2$ = event that cabinet 2 is chosen

Let $S$ = event that a silver coin is chosen

$P(C1|S) = \frac{P(S|C1)P(C1)}{(P(S|C1)P(C1)+P(S|C2)P(C2)}$

$P(C1|S) = \frac{1\times0.5}{1\times0.5 + 0.5\times0.5}$

$P(C1|S) = \frac{2}{3}$

However, I'm not entirely sure if this is correct, is it possible to get feedback/hints?

  • 2
    That looks right (aside from the typo that you wrote $P(S|C_1)$ rather than $P(C_1|S)$ on the LHS of your equations) – 2017-02-02
  • 1
    Well, I used Bayes' formula for this and according the theorem, they have to be reversed. – 2017-02-02
  • 1
    You should be computing $P(C_1|S)$. Bayes' formula is $P(C_1|S) = P(S|C_1)P(C_1)/P(S).$ Your RHS (and answer) is correct. It's just that the answer is $P(C_1|S) = 2/3$ – 2017-02-02
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    Oh, you are right, sorry! – 2017-02-02
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    The only reason why I'm doubting my answer is that I found different answers on websites such as (not my posts): http://answers.tutorvista.com/1081838/each-of-two-cabinets-identical-in-appearance-has-2-drawers-cabinet-a-contains-a-silver-coin-in-each-drawer-and-cabinet-b-contains-a-silver-coin-in-o.html – 2017-02-02
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    As well as http://www.transtutors.com/questions/each-of-2-cabinets-identical-in-appearance-has-2-drawers-cabinet-a-contains-a-silver-1652064.htm – 2017-02-02
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    And this: http://tinypic.com/view.php?pic=dzw4zk&s=9#.WJOj5xsrLIU – 2017-02-02
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    The first uses the same method as you but erroneously says $P(S|C_1) = .5$ rather than $1$ in the numerator. The second I can't read, and the third I can't understand. – 2017-02-02
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    The third basically asserts that $P(C_1\mid S) = P(C_1)$ through the erroneous assumption that if you have opened a drawer with a silver coin you are still equally likely to be in either of the two cabinets. Nope; wrong. You are equally likely to be in any of the three drawers containing a silver coin. – 2017-02-02

4 Answers 4

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IF you want a better understanding on how to solve this question, here is a link to the book that I used last semester in my probability math course. Head to page 141 and it should explain how conditional probability works. https://math.dartmouth.edu/~prob/prob/prob.pdf

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It looks good to me. You should make a tree diagram for these types of problems so you can easily compute the probabilities, assuming these events are mutually exclusive.

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Looks fine.   That is how you apply Bayes' Rule and the Law of Total Probability.

Another approach: You could have picked any drawer without bias, due to the drawer selection method.   So the silver coin you choose could have been equally likely to be any from the three silver coins, and two of these coins are in a cabinet containing another silver coin.   Thus verifying your result of $2/3$.

$\begin{array}{c} &&&\nearrow & \tfrac 12 & \circ \\ &\nearrow & \tfrac 12 & \searrow & \tfrac 12 & \circ \\[1ex] & \searrow & \tfrac 12 & \nearrow & \tfrac 12 & \circ \\ &&& \searrow & \tfrac 12 & \bullet \end{array}$

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Actually, after thinking about it hard, I don't think any of this is right. The question is asking the probability of a silver coin being in the other drawer, given that you already chose a silver coin.

I'm thinking of solving it as follows:

P(choosing silver coin|silver coin already chosen) $ = P(S|C1)P(C1) + P(S|C2)P(C2)$

$= (1)(0.5) + (0)(0.5) = \frac{1}{2}$

Since we don't know which cabinet we have chosen, we would have to calculate the total probability, because the probability that a silver coin is chosen really depends on which cabinet you have chosen. I think this reasoning makes more sense. What do you think?

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    No, your initial reasoning was right. The fact that you saw a silver coin (when you could have gotten a gold coin) tells you that you're more likely at the dresser that had two silver coins. – 2017-02-02
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    And indeed, $\mathsf P(S\mid C_1)\mathsf P(C_1)+\mathsf P(S\mid C_2)\mathsf P(C_2)~=~(1)(0.5)+(0.5)(0.5)~=~0.75~=~\mathsf P(S)$, the probability of selecting a silver coin. – 2017-02-02