# What is the probability of at least 2 blue balls given the probability of at least 1 blue ball?

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 $P(A)$?

• When we grab, is it with replacement or without? If without, I doubt you can find an exact answer without knowing how many balls there are. Commented Dec 10, 2012 at 7:18
• Oh! With replacement. Sorry about that. I must clarify. Commented Dec 10, 2012 at 7:21
• I think given what P(B) is, it's assumed that it's with replacement or from an infinite bag. Commented Dec 10, 2012 at 7:22

We need to assume that we are grabbing with replacement, which doesn't sound much like grabbing. The probability of $A$ is $1$ minus the probability of $0$ or $1$ blues.

You already found the probability of $0$ blues. For $1$ blue exactly, the probability is $\dbinom{n}{1}b(1-b)^{n-1}$. More generally, the probability of exactly $k$ blues is $\dbinom{n}{k}b^k(1-b)^{n-k}$ (binomial distribution).

Simply subtract both the probability that no balls are blue and the probability that exactly one ball is blue from 1.

$P(A) = 1 - (1-b)^n - n b^1 (1-b)^{n-1}$

• Thanks, why is the probability that exactly one ball is blue $b(1-b)^{n-1}$? Couldn't any one of the $n$ balls be blue, resulting in $nb(1-b)^{n-1}$? Commented Dec 10, 2012 at 7:23
• Oh, sorry, I meant to type that n. Thanks. Commented Dec 10, 2012 at 7:29