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Question:

Amy has a bowl of 5 red cherries and 8 purple cherries. She takes out cherries one at a time until there are no red cherries left. What is the probability that there are exactly 2 cherries left in the bowl?

Solution attempt: I imagine that the $13$ cherries are arranged in a random sequence $X_1\ldots X_{13}$, and Amy picks them in that order. There are $13!$ possible orderings of this sequence. Out of these orderings, the ones for which exactly $2$ cherries are left in a bowl have the form $X_1\ldots X_{10}{\rm RPP}$, i.e. the sequence ends with a red cherry followed by two purple cherries. There are $5$ ways to select the red cherry, ${8\choose 2}$ ways to pick the two purple cherries, and $10!$ ways to arrange the remaining $10$ cherries. The probability is then $$\frac{5\times{8\choose 2}\times 10!}{13!} = \frac{35}{429} \approx .082$$ I'm solving this on a website, and it does not accept my answer. I ran a Monte Carlo simulation as a sanity check, and it gave me values around $.1$, so it seems I am indeed lowballing. What am I missing? Thanks!

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    $\begingroup$ And the ways to arrange the $2$ purple cherries left in the bowl? $\endgroup$
    – peterwhy
    Commented Jul 31, 2023 at 2:41
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    $\begingroup$ @peterwhy Ope! That'll do it :) feel free to post this as an answer and I'll accept it. $\endgroup$ Commented Jul 31, 2023 at 2:45
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    $\begingroup$ The answer depends on whether Amy is taking the cherries out randomly or is cherry picking her preferred color. $\endgroup$
    – user10478
    Commented Jul 31, 2023 at 4:39
  • $\begingroup$ Based on the answers below, your Monte Carlo simulation is probably buggy. A good simulation run say a million times should have gotten you to the vicinity of 0.163 rather than 0.1. You might then have noticed that it seems that you were off by a factor of 2. $\endgroup$ Commented Jul 31, 2023 at 13:44
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    $\begingroup$ @JohnColeman yeah, I was sloppy all around, eh? I miscalculated the probability of selecting red vs. purple at each choice! $\endgroup$ Commented Jul 31, 2023 at 14:11

3 Answers 3

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The last $3$ picked (in order) must be red, purple, purple.

The simplest solution is to see that by symmetry, P(last $3\; RPP$) = P(first $3\;PPR$), and hence $Pr = \frac8{13}\frac7{12}\frac5{11} = \frac{70}{429} \approx 0.16317$.

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  • $\begingroup$ Surprising to see two downvotes, too ! $\endgroup$ Commented Aug 1, 2023 at 20:04
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The approach in the question considers all $13!$ permutations of different cherries, but only misses the $2$ ways to order the two purple cherries left in the bowl. The probability is then

$$\frac{5\times{8\choose 2}\times 2\times 10!}{13!} = \frac{70}{429} \approx .163$$

Alternatively, I consider the sequences of $5$ identical red cherries and $8$ identical purple cherries. There are $\binom {13}5$ possible sequences.

Out of these sequences, the ones for which exactly $2$ cherries are left in a bow have the form $X_1\ldots X_{10}RPP$. Choose $4$ positions among $X_1,\ldots,X_{10}$ for the red cherries. The probability is then

$$\frac{\binom{10}{4}}{\binom{13}{5}} = \frac{\frac{10!}{4!6!}\times5!8!}{13!} = \frac{\frac{5!}{4!}\times\frac{8!}{6!2!}\times 2\times10!}{13!} = \frac{70}{429} \approx .163$$

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Alternative approach:

In order for there to be exactly 2 (purple) cherries left, two things have to happen:

  • The first 10 cherries picked must contain exactly $~4~$ red cherries and $~6~$ purple cherries.
    The probability of this happening is
    $~\displaystyle \frac{\binom{5}{4} \times \binom{8}{6}}{\binom{13}{10}} = \frac{70}{143}.$

  • With $~3~$ cherries left, one of which is red, the $~11$'th cherry picked must be red.
    The probability of this happening is $~\dfrac{1}{3}.$

final answer: $~\displaystyle \dfrac{70}{143} \times \dfrac{1}{3} = \dfrac{70}{429}.$

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