Ok since I couldn't get in here I was surfing the net for

Question

Ok since I couldn't get in here I was surfing the net for some questions. Here's one that I'm a little confused.

There are R red balls and W white balls. You pick a ball, discard it. And then you pick another ball, discard it, and so on. What's the probability of getting a white ball in the kth pick?

How would you do this kind of question?

HongHu · Answer

Any thought on this one?

prep_gmat · Answer

for k > 1:
(k-1)RCk-1 * W / k * (R+W) Ck

for k = 1
w/r+w

probably there is a way to combine the two as well...

HongHu · Answer

How about let's put in some numbers. Say 5 white balls, 3 red balls. What would be the probability of picking a red ball in the second pick?

Total pick: P(8,2)=56
If the first pick is a red: C(3,1)C(2,1)=6
If the first pick is a white: C(5,1)C(3,1)=15
Probability = 21/56=3/8

Now what if k=3? k=4? It's going to be very hard to do it case by case. There should be some general ways to do it.

anirban16 · Answer

Hi Hong Hu,
I think this is how to approach the problem:

R red balls and W white balls. 
Probability of picking W on the kth pick ==> Till k-1 picks you only picked R balls and only on the kth pick you picked the white ball. 

since this is a general problem without involving any numbers the answer will have to be in a series like expression.
Now lets give index to our k.

Say k = 1 ==> On the very first pick you get W

P(k) = P(1) = W/(R+W)

Say k=2 

P(k) = P(2)= R/(R+W)*W/{(R-1)+W}

Say k=3

P(k) = P(3) = R/R+W*(R-1)/{(R-1)+W}*W/{(R-2)+W}

For k > 3

P(k) = R(R-1)..(R-(k-2))W/(R+W)(R-1+W)(R-2+W)..(R-(k-1)+W) And this is a product series.

Anirban

HongHu · Answer

The flaw in your reasoning is that you are assuming that until the kth pick you only get red balls, and only on the kth pick you get a white ball. That is not a correct assumption. You could get Red or White balls for the previous picks.

anirban16 · Answer

Sorry.
I also didn't read the word discard. But we can proceed with the same logic only the series will get bigger

baggarwal · Answer

In the original stem it said the white ball should be selected in kth pick. Hong your answer refers to a red ball picked in the kth pick. So the 5/8 is the likelihood of getting a white ball. 

The general equation will likely be W/W+R. I think the answer for all K will be 5/8.

Ok since I couldn't get in here I was surfing the net for

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