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TranscendArcu
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Homework Statement
Homework Equations
The Attempt at a Solution
I've done part b) so I only want to talk about part a). So I said, let Ei be the event that I do not get the kth players card. It seems logical that the union of all the E-complements should be one.
Theorem 3.8 clearly makes use of intersections, so I'll compute for any k failures to get pictures [itex]P(E_1 \cap E_2 \cap ... \cap E_k) = (\frac{n-k}{n})^m[/itex] given that I buy m boxes of cereal. I have [itex]_n C_k[/itex] ways of failing to pick some k number of pictures. Thus, following the " subtract the probabilities of all possible two-way intersections, add the probability of all three-way intersections"-principle of Theorem 3.8, I can write as the k's vary:
[itex]1 - (_n C _1)(\frac{n-1}{n})^m + (_n C _2)(\frac{n-2}{n})^m - (_n C _3)(\frac{n-3}{n})^m + ... + (-1)^{n-1}(_n C _{n-1})(\frac{1}{n})^m[/itex]
Note that the nth term would go to zero.
So how does it look?