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'There are 100 prizes, with one worth $1, one worth 2, . . . , and one worth 100. There are 100 boxes, each of which contains one of the prizes. You get 5 prizes by picking random boxes one at a time, without replacement. Find the PMF of how much your most valuable prize is worth (as a simple expression in terms of binomial coefficients)'.--Introduction to Probability(Joseph K. Blitzstein, Jessica Hwang)

From my point of view the probability is equal to:

$P(X=k)$=$\binom{5}{1}$/$\binom{100}{5}$ for k=1,2,...100

Because there are in total $C(100,5)$ ways of choosing the box and $C(5,1)$ of choosing the most valuable prize.

What's wrong with this reasoning?

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You are searching for probability that $k$ is the most valuable prize you chose. That is probability that all prizes you chose are less valuable than $k+1$ and that you chose $k$, which means that not all prizes are less than $k$.

Denote by $Y_i$ event that all 5 prizes you chose are less than $i$.

Then: $$P(Y_i)=\frac{i \choose 5}{100 \choose 5}$$ and: $$P(X=k)=P(Y_{k+1}\cap \neg Y_k)=P(Y_{k+1})-P(Y_{k+1} \cap Y_k)=P(Y_{k+1})-P(Y_k)=\frac{{{k}\choose{5}}-{{k-1}\choose{5}}}{{100}\choose{5}}$$

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  • $\begingroup$ @M C to be honest I can't understand your solution, why all prizes must be less valuable than k+1? For example, my most valuable prize is 5, then you suggest that all others must be less valuable than 6. Well, it is true but why you specify k+1 and not k. And also what does it mean $\binom{k}{5}$ choose from k most valuable 5? why? However, you gave me a hint I decided to fix the most valuable prize and choose the other 4 from remaining. That gives $\binom{k-1}{4}$/$\binom{100}{5}$ for k=5,6,7...100 $\endgroup$ – Sargis Iskandaryan Nov 5 '17 at 21:32
  • $\begingroup$ Here I meant that all, including most valuable one must be less than $k+1$ ($6$), but then it must not be true, that all are less than $k$ ($5$), $k \choose 5$ is number of ways to choose $5$ prizes, all less valuable than $k+1$ $\endgroup$ – M C Nov 5 '17 at 21:44

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