Expected number of red balls drawn before drawing a white ball

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Homework Statement



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?

Homework Equations



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

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

This is so difficult to sum!
 
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.
 
But the answer is 46/5 ...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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