Question:
There are three identicalboxes, each provides with two drawers. In the first, each drawer contains a gold coin, in the third, each drawer contains a silver coin and in the second, one drawer contains a gold and the other a silver coin. A box is selected at random and one of the drawers is opened. If a gold coin is found, what is the probabilitythat the box chosen is the second one?
My try:
Let, $P(A)\Rightarrow$ Probability of choosing first box
$P(B)\Rightarrow$ Probability of choosing second box
$P(C)\Rightarrow$ Probability of choosing third box
$G\Rightarrow$ is the event of finding a gold coin
$S\Rightarrow$ is the event of finding a silver coin
Now, $P(A)= P(B)= P(C) = \frac 13$
And we have to find $P(B \mid G)$.
Now, according to the formula,
$$P(B \mid G) = \frac{P(B \cap G)}{P(G)}$$
But, I can't understand how to find $P(G)$ and $P(B \cap G)$. Please help.
This is a problem that requires you to know Bayes' theorem on conditional probability. It is just basically an extension of what you have written out. What it says is, $$\ P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$ You can try it to prove it yourself, from what you have written itself.
Now, to the problem, it asks us to find the probability that the second chest is selected given that a gold coin is obtained. So, we are asked to find $\ P(B|G)$. From Bayes' theorem, it is, $$\ P(B|G)=\frac{P(G|B)P(B)}{P(G)}$$ Can you proceed?
Anyway, I continue. We write it using Bayes' form because we can find out each term in the RHS. Now, $$\ \frac{P(G|B)P(B)}{P(G)}=\frac{\frac{1}{2} \cdot\frac{1}{3}}{\frac{1}{2} \cdot\frac{1}{3}+1 \cdot \frac{1}{3}}=\frac{1}{3}$$ I hope the answer is correct and hope it helps!