# Multi Conditional Probability of selecting balls

My question is this; what is the probability of withdrawing at least one red ball and at least one blue ball from a bag with a certain number of attempts?

The requirements are, this bag has n number of balls, consisting of x-red balls, y-blue balls and z-green balls. The balls are placed back into the bag after being removed so the probability of any colour being withdrawn is constant.

In order to achieve a win at least one blue and at least one red ball must be withdrawn, e.g. if you only decide to make 3 withdrawals, then a win is one red ball, one blue ball and one ball of any colour.

My main question is, is there a formula that can be constructed to answer this question? If so what is that formula?

## 1 Answer

We first find the probability of failure, that is, the probability that we get no red, or no blue (this includes the possibility we get neither a red nor a blue). Once we have that, the probability of success is easy to find. Let us suppose that we do the experiment $k$ times.

Let $A$ be the event we get no red in $k$ trials, and $B$ be the event we get no blue. We want $\Pr(A\cup B)$. Recall that in general $$\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B).\tag{1}$$ We find the probabilities on the right of (1).

The probability of no red in $k$ trials is $\left(\frac{n-x}{n}\right)^k$. The probability of no blue is $\left(\frac{n-y}{n}\right)^k$. And the probability of no red and no blue is $\left(\frac{n-x-y}{n}\right)^k$.

Now put the pieces together.

• Thank you very much, the annoying thing was I answered this question myself about 2 years ago but I forgot the method I used to answer it. – Bodmas12 Aug 8 '14 at 15:09
• You are welcome. It is annoying. However, things you once knew can be recovered fairly quickly. But it's not instantaneous. – André Nicolas Aug 8 '14 at 15:12