There is a box with four drawers. The contents are respectively $GGG, GGS, GSS$, and $SSS$. You pick a drawer at random and a coin at random. The coin is $G$. What is the probability that there is another $G$ coin in the drawer?
Let $$A:= \text{The first coin that i take from the drawer is } G. \\ B:= \text{The second coin that i take from the drawer is G} $$
So i need to find $P(B \mid A)$. \begin{align} P(B \mid A) &= \frac{P(B \cap A)}{P(A)} \\ &= \frac{\frac{1}{4}\cdot1\cdot1+\frac{1}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}}{\frac{1}{4}\cdot1+\frac{1}{4}\cdot\frac{2}{3}+\frac{1}{4}\cdot\frac{1}{3}} \\ P(B \mid A) &= \frac{2}{3} \end{align}
So the probability that there is another gold coin in the drawer is $\frac{2}{3}$.
Am i right?, the book says that the answer is $\frac{5}{6}$
You calculated precisely what you said you’d calculate, with $B$ the event that the second coin that you take from the drawer is a $G$. But that wasn’t the question. The question was about the probability that there’s another $G$ coin in the drawer. Fortunately you only have to change one of the numbers in your calculation to fix the mistake.