Probability with and without replacement

[mutzy188]

Homework Statement

An urn contains five balls, one marked WIN and four marked LOSE. You and another player take turns selecting a ball from the urn, one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done

(a) with replacement - answer = 5/9
(b) without replacement - answer = 3/5

The Attempt at a Solution

(a) P(W) + P(LW) + P(LLW) + . . . . .

(1/5) + (4/5)(1/5) + (4/5)(4/5)(1/5) . . . .

so its is a series

SUM (1/5)(4/5)^(n-1) starting at n=1 to infinity
when I did this I got 1 and not 5/9

(b) P(W) + P(LW) + P(LLW) . . .

(1/5) + (4/5)(1/4) + (4/5)(3/4)(1/3) . . ..

SUM (1/5)( . . . and I got stuck here

Any Help would be appreciated

Thanks

 

[learningphysics]

You have not considered that the second person is also taking out a ball... you did the problem as if it was just one person...

 

[EnumaElish]

For me (b) is easier: Let P(L|x) be the probability of "L" when there are x balls in the urn. P(W|x) = 1 - P(L|x).

P(W|5) + P(L|5)P(L|4)P(W|3) + the 3rd term = 1/5 + 4/5 3/4 1/3 + the 3rd term.

Can you guess what the 3rd term is?

 

[mutzy188]

For me (b) is easier: Let P(L|x) be the probability of "L" when there are x balls in the urn. P(W|x) = 1 - P(L|x).

P(W|5) + P(L|5)P(L|4)P(W|3) + the 3rd term = 1/5 + 4/5 3/4 1/3 + the 3rd term.

Can you guess what the 3rd term is?

The third term should be:

P(L|5)P(L|4)P(L|3)P(W|2) = 1/5

 

[learningphysics]

The third term should be:

P(L|5)P(L|4)P(L|3)P(W|2) = 1/5

Although the third term does come out to 1/5... just wanted to point out it should be:

P(L|5)P(L|4)P(L|3)P(L|2)P(W|1) = 1/5

because the expression P(L|5)P(L|4)P(L|3)P(W|2) means the second guy wins not you...

 

[EnumaElish]

mutzy188, do you get 3/5 for b?

 

[mutzy188]

mutzy188, do you get 3/5 for b?

Yes I did. Thank you very much for your help