Expected tries to to remove ball from urn

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SUMMARY

The discussion centers on calculating the expected number of attempts to draw a black ball from an urn containing 3 red and 1 black ball, specifically without replacement. The initial approach utilized the expected value formula E[X]=1*P(X=1) + 2*P(X=2) + ..., but the user struggled with applying the hypergeometric distribution correctly. A simpler method was proposed, emphasizing that the probability of drawing the black ball remains 1/4 for the first attempt, but changes for subsequent draws due to the lack of replacement. The final solution involves calculating probabilities for each possible outcome, leading to a clearer understanding of the expected value.

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funnyguy
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I have a problem dealing with expected values. I'll jump right into the problem:

There are 3 red and 1 black balls in an urn. What is the expected number of times a ball must be removed before a black ball is removed.

I originally thought of this using

E[X]=1*P(X=1) + 2*P(X=2) + ...

I filled in P's using hyper geometric distribution. But didn't get the correct result.

Trying a uniform for each trial I did this
1*1/4 + 2*1/3 + 3*1/2 + 4*1/1

I found the following as a solution from a LONG derivation of formulas:
k(r+b+1)/(b+1) where k=1 in this case.
From a problem setup the same but using negative hyper geometric.

This makes no sense to me. Is there some more simple way of thinking of this?
 
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The probability that the ball is picked in the nth attempt is just 1/4, as all balls are equal. This allows to calculate the expected value easily.

Trying a uniform for each trial I did this
1*1/4 + 2*1/3 + 3*1/2 + 4*1/1
You would have to consider the probability that the ball was not drawn before in each summand.
 
mfb said:
The probability that the ball is picked in the nth attempt is just 1/4, as all balls are equal. This allows to calculate the expected value easily.You would have to consider the probability that the ball was not drawn before in each summand.

Sorry, but I forgot to mention this is without replacement.

So to start the black ball probability is 1/4, but after the 1st attempt the black ball is 1 of 3 remaining, etc.
 
funnyguy said:
Sorry, but I forgot to mention this is without replacement.
I know.
So to start the black ball probability is 1/4, but after the 1st attempt the black ball is 1 of 3 remaining, etc.
Sure, but the probability that you draw a second ball is just 3/4. If you multiply 1/3 by 3/4, you get 1/4 again.
 
mfb said:
I know.
Sure, but the probability that you draw a second ball is just 3/4. If you multiply 1/3 by 3/4, you get 1/4 again.

Maybe I see what you're saying.

P(X=1) = 1/4
P(X=2) implies P(X=2) AND P(X!=1) = 1/3 * (1- P(X=1))

?
 
Right.
It is possible to extend this for P(X=3) and P(X=4), but that is unhandy. You can get the probabilities in a direct way as well.
 
b, rb, rrb and rrrb are the 4 possible cases corresponding to your X=0,1,2,3. Find the later three probabilities and hence the expectation.
 

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