Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $P(A)$?

share|cite|improve this question
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. – André Nicolas Dec 10 '12 at 7:18
Oh! With replacement. Sorry about that. I must clarify. – John Hoffman Dec 10 '12 at 7:21
I think given what P(B) is, it's assumed that it's with replacement or from an infinite bag. – Joe Z. Dec 10 '12 at 7:22
up vote 1 down vote accepted

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).

share|cite|improve this answer

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

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

share|cite|improve this answer
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 $nb(1-b)^{n-1}$? – John Hoffman Dec 10 '12 at 7:23
Oh, sorry, I meant to type that n. Thanks. – Joe Z. Dec 10 '12 at 7:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.