Expected number of red balls drawn before drawing a white ball

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Homework Help Overview

The problem involves calculating the expected number of red balls drawn from a bag containing 50 balls, of which 4 are white and 46 are red, before drawing a white ball. The context is within probability and discrete random variables.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the probability of drawing red balls and attempt to express this in terms of a sum. Questions arise regarding the formulation of the probability P(X ≥ k) and whether the final answer can be represented in a simpler form.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to summing probabilities and questioning the nature of the final expression. Some participants are seeking hints while others are attempting to derive a closed form for the expected value.

Contextual Notes

There is a mention of the challenge in summing the probabilities and the potential for the answer to remain in a summation form. The original poster expresses difficulty in finding the necessary probabilities.

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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 ...
 

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