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Q: an urn contains 2 red and 3 black balls. Players 1 and 2 withdraw balls from the urn consecutively without replacement until the second red ball is selected. player 1 draws first, then player 2, and so on. Find the probability that player 1 selects the second red ball

My approach: if i let P_1r be player 1 pick red, P_2b = player 2 pick black so on.

P(P_1r) = 2/5

P(P_1b) = 3/5

P(P_2rlP_1r) = 1/4...

but this does not even go near to the right answer... I'm sure there is easy and efficient way to approach this kind of question but i'm stuck with it. can anyone help? thanks in advance!

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    $\begingroup$ HInt: There are general ways to do this that work better when there are lots of cases, but here I'd say: just enumerate. There are only $\binom 52 =10$ ways to arrange the balls in order of selection, all equally likely. Just list them. $\endgroup$
    – lulu
    Dec 6, 2015 at 19:23

2 Answers 2

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Let the string of Rs and Xs denote the color of balls (red and black, respectively) drawn in the order they were drawn, where the leftmost letter represents the first ball drawn, and the rightmost is the last ball drawn (e.g. $RXR$ means 1 drew red, 2 drew blue, then 1 drew red).

We are looking for when the first player draws the second red ball. Notice the first player only draws the first, third, or fifth time, and the first draw cannot be the second red ball. So we're looking for R in the third slot as the second R to appear, or R in the fifth slot as the second R to appear.

In order words, $RXRXX$, $XRRXX$, $RXXXR$, $XRXXR$, $XXRXR$, or $XXXRR$. As lulu noted, the outcomes are equally probable and there are $5\choose 2$ outcomes.

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  • $\begingroup$ RXRXX will have $\frac{2}{5}$ x $\frac{3}{4}$ x $\frac{1}{3}$ and that is $\frac{1}{10}$ and there are 6 cases so its $\frac{6}{10}$ ? $\endgroup$
    – Allie
    Dec 6, 2015 at 20:07
  • $\begingroup$ Yes, that's what I got. $\endgroup$
    – manofbear
    Dec 6, 2015 at 20:12
  • $\begingroup$ awesome! thanks $\endgroup$
    – Allie
    Dec 6, 2015 at 20:16
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You have rather small numbers, so you can go case by case. If P1 starts, what you need is the probability to sample a red ball on either 1st, 3rd of 5th trial. In the first case it's $\frac{2}{5}$, in the second it's $ 2 \cdot\frac{2}{5} \cdot \frac{3}{4} + \frac{3}{5} \cdot \frac{2}{4}$. Can you do the last one?

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  • $\begingroup$ last one would be 2x $\frac{2}{5}$ x $\frac{3}{4}$ x $\frac{2}{3}$ x $\frac{1}{2}$ + 2x $\frac{3}{5}$ x $\frac{2}{4}$ x $\frac{2}{3}$ x $\frac{1}{2}$? $\endgroup$
    – Allie
    Dec 6, 2015 at 20:15
  • $\begingroup$ $\frac{\binom{4}{1}}{\binom{5}{2}}$ $\endgroup$
    – Alex
    Dec 6, 2015 at 20:44
  • $\begingroup$ 4/10 so same answer! okay so do you add first, second, and third case to get the final answer? 2/5 + 9/10 + 4/10? but that exceeds 1 $\endgroup$
    – Allie
    Dec 6, 2015 at 21:00
  • $\begingroup$ Apologies, I didn't read too carefully: 'the second red ball'. Remove the first case; in the second case (3 trials) remove one of the cases, with just 1 red ball $\endgroup$
    – Alex
    Dec 6, 2015 at 21:39

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