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John goes to the grocery. His mother sent him to buy $20$ peaches and requested him to be sure that the peaches were mature. Suppose the probability of a peach of being mature is $p$ and suppose that John picks the $20$ peaches at random without verifying the maturity of each peach. His mother needs $3$ peaches to make a fruit salad and, as she doesn't trust her son, she checks each peach at home. If one peach is mature, she keeps it and if not, she throws it away. She stops to check each peach she grabs when she gets the $3$ peaches she needs for the salad. What is the probability that she has to throw away exactly $5$ peaches?

I am not sure if I've answered this question correctly so I'll write my solution.

If a success is picking a mature peach, then we can consider the negative binomial random variable $X \sim NB(3,p)$ that counts the number of trials in order to obtain $3$ successes. So, the probability of the problem is $$P(X=8)={7 \choose 2}p^3(1-p)^5$$

Is this solution correct? I get confused by the fact that we have $20$ peaches to choose.

So,

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    $\begingroup$ As always, phrasing in these problems can be ambiguous. You have assumed that mom is to throw out $exactly$ five peaches, but the problem could just as easily be read as $at\;least$ five peaches. $\endgroup$ – lulu Apr 28 '16 at 1:26
  • $\begingroup$ @lulu makes a really good point that I hadn't even though of. I think it has to be read as $at least$ five peaches, since if she has to throw away six peaches, she also threw away five. $\endgroup$ – Chill2Macht Apr 28 '16 at 1:34
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    $\begingroup$ It seems like your answer is correct under your assumption, although I might sum $P(X=8)+...+P(X=20)$ and then also account for the cases where only $2, 1,$ or $0$ peaches are mature. $\endgroup$ – Chill2Macht Apr 28 '16 at 1:35
  • $\begingroup$ Sorry for the ambiguity, I was meaning to say exactly five. $\endgroup$ – user156441 Apr 28 '16 at 1:42

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