Expected tries to to remove ball from urn

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In summary, the expected number of times a ball must be removed before a black ball is removed is given by the formula k(r+b+1)/(b+1), where k=1 and r and b are the number of red and black balls, respectively. This can be derived using the negative hyper geometric distribution. However, a simpler way to calculate this is by considering the probability that the ball is picked in the nth attempt, which is just 1/4 for all attempts due to equal probabilities for all balls. This can be adjusted for the fact that this is without replacement by taking into account the probability that the ball was not drawn before in each attempt.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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))

?
 
  • #6
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.
 
  • #7
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.
 

Related to Expected tries to to remove ball from urn

1. What is an urn in the context of this question?

An urn is a container that holds a certain number of balls. In the context of this question, the urn is used as a model for understanding probability and the likelihood of selecting a specific ball from a set of balls.

2. What does "expected tries" mean?

Expected tries refers to the average number of attempts it would take to select a specific ball from an urn. This value is calculated by taking into account the total number of balls in the urn and the probability of selecting the desired ball on each try.

3. How is the expected number of tries calculated?

The expected number of tries is calculated by dividing the total number of balls in the urn by the probability of selecting the desired ball on each try. For example, if there are 10 balls in the urn and the probability of selecting the desired ball on each try is 1/10, then the expected number of tries would be 10/1 = 10.

4. Why is the expected number of tries important?

The expected number of tries is important because it helps us understand the likelihood of selecting a specific ball from an urn. It also allows us to make predictions and calculate probabilities in situations where multiple tries are involved.

5. Can the expected number of tries be less than 1?

No, the expected number of tries cannot be less than 1. This is because it represents the average number of attempts it would take to select a specific ball from an urn, and it is not possible to have a fraction of an attempt. However, the expected number of tries can be a decimal value, indicating that it may take slightly more than one try on average to select the desired ball.

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