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A desk has three drawers. The first contains two gold coins, the second has two silver coins and the third has one gold and one silver coin. A drawer is selected at random and a coin is drawn at random from the drawer. Suppose that the coin selected was silver. Use Bayes's Theorem to find the probability that the other coin in that drawer is gold.

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  • $\begingroup$ What have you done so far? Do you know what Bayes' Theorem looks like? $\endgroup$ – Emily Jul 22 '13 at 13:40
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    $\begingroup$ This problem can be solved using the definition of conditional probability: $P(B\mid A) \cdot P(A) = P(B \cap A)$. Bayes's Theorem is useful if you need to "flip" the condition, but doing that here does not really change the problem... $\endgroup$ – angryavian Jul 22 '13 at 13:46
  • $\begingroup$ This is what I have so far: I have to find P(G|S). P(G) = 2/3 and P(S) = 1/3. P(S|G)= P(S&G)/P(G) = (2/6)/(2/3) = 0.5. Using Bayes's formula I get P(G|S) = P(G)*P(S|G)/P(G)*P(G)*P(S|G) + P(S)*P(S|G) = 2/3. Is this the right answer? $\endgroup$ – OGC Jul 22 '13 at 13:53
  • $\begingroup$ What exactly do the events $S$ and $G$ represent? I'm guessing $S$ is the event that the coin you selected was silver. How do you get $P(S)=\frac 13$? Likewise, if you mean that $G$ represents that the event that the other coin in the drawer is gold, how do you get $P(G) = \frac 23$? $\endgroup$ – angryavian Jul 22 '13 at 14:07
  • $\begingroup$ G is the event that the coin selected is gold and S is the event that the coin selected is silver. P(G) is the probability of a gold coin being selected from the three drawers combined. So, P(G) = 4/6 = 2/3 and same concept for P(S) = 2/6 = 1/3. $\endgroup$ – OGC Jul 22 '13 at 14:11
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You select a drawer and select a coin from that drawer.

Let $A$ denote the event that the coin that you select is silver.

Let $B$ denote the event that the other coin in the drawer is gold.

You are equally likely to pick any coin, so

$$P(A) = \frac{3\text{ silver coins}}{6 \text{ coins}} = \frac 12$$

There are 6 different ways you could have selected a coin. Only one of them results in you selecting a silver coin AND the other coin in the drawer being gold, so $$P(A \cap B) = \frac{1\text{ desired outcome}}{6 \text{ total outcomes}} = \frac 16$$

By the definition of conditional probability, $$P(B \mid A) = \frac{P(B \cap A)}{P(A)} = \frac{1/6}{1/2} = \frac 13$$


As I mentioned in my comment, Bayes's Theorem is useful if you want to know $P(B\mid A)$ but you only know $P(A\mid B)$. Here, we don't know $P(A \mid B)$ either, so Bayes's Theorem doesn't really help.

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  • $\begingroup$ I see. I am a bit confused about the question saying "Use Bayes's Theorem to find the probability that the other coin in that drawer is gold". I think that we only have to use Bayes's Thm to solve this problem as the question asks for it. Even though your reasoning makes perfect sense. $\endgroup$ – OGC Jul 22 '13 at 14:46
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Let the choice of picking a drawer amongst 3 represent 3 events called $gg, gs, ss$ for both gold, one gold one silver and 2 silver drawer respectively. These probabilities are $\frac{1}{3}$ each.

From the $2g$ drawer, probability of picking a silver coin $p(s|gg) = 0$.

Similarly, $p(s|gs) = \frac{1}{2}$ and $p(s|ss) = 1$.

The only way the other coin can be gold is if you've picked the $gs$ drawer. So you're trying to find $p(gs|s)$.

By Bayes' theorem,

$$p(gs|s) = \frac{p(s|gs)\cdot p(gs)}{p(s|gs)\cdot p(gs) + p(s|gg)\cdot p(gg) + p(s|ss)\cdot p(ss)} \\ = \frac{\frac{1}{2}\cdot\frac{1}{3}}{\frac{1}{2}\cdot\frac{1}{3} + 0 \cdot \frac{1}{3} + 1\cdot\frac{1}{3}} \\ = \frac{1}{3}$$

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