# 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 The balls are placed back into the bag after being removed so the probability of any colour being withdrawn is constant.

In order to achieve a win at least one blue and at least one red ball must be withdrawn, e.g. if you only decide to make 3 withdrawals, then a win is one red ball, one blue ball and one ball of any colour.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

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