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An urn contains 2 blue and 3 red balls. Alice and David withdraw balls from the urn consecutively without replacement until the second blue ball is selected. Alice draws first, then David, and so on. Find the probability that Alice selects the second blue ball.

I've tried to solve by finding the possible ways Alice selects the second blue ball, i.e. the third or fifth ball is blue. So we have: brb, rbb, rrrbb, rrbrb, rbrrb, brrrb.

Not sure what to do from here. I was thinking about finding the probability for each outcome but would there be a better way to do this?

Thanks in advance!

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    $\begingroup$ For heaven's sake, there are only $\binom 52=10$ possible orderings. Just write them all down. $\endgroup$ – lulu Oct 22 '17 at 17:46
  • $\begingroup$ Why is it (5 choose 2)? I did get 10 by writing it out but wasn't sure how to calculate it. $\endgroup$ – Lin Oct 22 '17 at 18:07
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    $\begingroup$ Because there are $2$ blue balls out of $5$. $\endgroup$ – lulu Oct 22 '17 at 18:08
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There are only $10$ possible orderings and these are equi-probable. They are $$brrrb,brrbr,brbrr,bbrrr,rbrrb,rbrbr, rbbrr,rrbrb,rrrbbr,rrrbb$$ and it is easy to see that of these only numbers $1,3,5,7,8,10$ pass your test so the answer is $$\frac 6{10}=\boxed {\frac 35}$$

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  • $\begingroup$ But how would you solve it if the number is larger so I can't just write out all possible outcomes? $\endgroup$ – Lin Oct 22 '17 at 18:19
  • $\begingroup$ I would solve it recursively. If the first one is blue, the answer is $\frac 12$. If the first one is red, then you are down to one fewer (with positions switched). $\endgroup$ – lulu Oct 22 '17 at 18:20
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The probability that the second blue ball is selected third is $\frac{2}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}$+ $\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{1}{3} = \frac{1}{5}$.

The probability that the second blue ball is selected fifth is the same as the probability of a blue ball being selected first, since we know that if blue is last, then the first blue ball must already have been selected. Therefore the probability is $\frac{2}{5}$

Therefore the total probability that Alice will select the second blue ball is $\frac{3}{5}$

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As I got a different result I told my friend Ruby about the game and asked her to run a few and tell me her results. She got:

n=1000000 alice wins=600482 => p=0.600482

which agrees with the other results. (Sigh)

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