# Expected number of red balls drawn before drawing a white ball

1. Jun 23, 2010

### txy

1. The problem statement, all variables and given/known data

There are 50 balls in a bag, all are the same except for the color. 4 of them are white while the rest (46) of them are red. Now randomly draw balls from the bag, one at a time, without replacement. What is the expected number of red balls that you will draw before you first draw a white ball?

2. Relevant equations

If a discrete random variable X takes only positive values, then its expectation
$$E(X) = \sum_{k=1}^{+\infty}P(X \geq k)$$

3. The attempt at a solution
Let X be the number of red balls drawn before first drawing a white ball. I can't seem to find $$P(X \geq k)$$. Can somebody give me a hint?

2. Jun 23, 2010

### jeppetrost

Say it out loud :)
What is the probability, that you draw 1 or more red balls? It is of course the probability of the first ball being red.
What is the probability, that you draw 2 or more red balls?
Etc.
You should get a nice sum, which you will be able to write in a closed form.

3. Jun 24, 2010

### txy

The probability of the first k balls is red is $$\frac{C_{46}^{k}}{C_{50}^{k}}$$

This is so difficult to sum!

4. Jun 24, 2010

### tmccullough

Is it OK for the answer to just be a sum? It might be the case that this doesn't have a better looking form than a sum.

5. Jun 25, 2010

### txy

But the answer is 46/5 ...