A bag has red, blue, and green balls. The probabilities of randomly grabbing a red, blue, and green ball from the bag (with replacement) are $r$, $b$, and $g$ respectively. I randomly grab $n$ balls from the bag. What's the probability that at least 2 out of the $n$ balls are blue given that one of them is blue?
Here is what I have tried.
Let $ A \rightarrow\text{ The event in which at least 2 out of the $n$ balls are blue.} \\ B \rightarrow\text{ The event in which at least 1 out of the $n$ balls is blue.} $
$ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} \\ P(B) = 1 - (1 - b)^n $
As you see, I got $P(B)$ from subtracting out a complement probability. However, how do I get $P(A)$?