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Suppose ๐‘ต different toys are offered in boxes of a favorite brand of cereal. You want to collect all the different toys. How many boxes of cereal do you expect to have to buy to get all ๐‘ต toys? Let the random variable ๐‘ฟ be the number of boxes of cereal you need to buy to get each toy at least once. Also, assume it is equally likely that any one of the ๐‘ต toys will be in each box of cereal.

a. Show that $๐‘ฟ = ๐‘ฟ_๐ŸŽ + ๐‘ฟ_๐Ÿ + ๐‘ฟ_๐Ÿ + โ‹ฏ + ๐‘ฟ_{๐‘ตโˆ’๐Ÿ}$ , where $๐‘ฟ_๐’Œ$ has a geometric distribution with probability $\frac{๐‘ตโˆ’๐’Œ}{๐‘ต}$.

Hint: consider the random variable which is the number of boxes needed to get the ๐’๐’•๐’‰ toy after getting ๐’ โˆ’ ๐Ÿ of all of them.

b. Find ๐‘ฌ(๐‘ฟ). The correct answer indicates that to get all eight toys offered with the current MacDonalds Happy Meal, you would, under the assumptions of the calculation, expect to buy about $22$ Happy Meals.

I was able to calculate part b as $E(X) = N(\frac{1}{N} + \frac{1}{N-1} + ... + \frac{1}{N - (N-1)})$ but I can't figure out how to start part a.

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    $\begingroup$ There is literature on this. It's called the Coupon Collector's Problem. $\endgroup$ – Gerry Myerson Feb 12 at 2:58
  • $\begingroup$ I found information on the coupon collector's problem before, but I haven't been able to properly link it to what part a of the question is asking me to do. I was able to use it to solve part b, but whatever material I read on the problem doesn't clear up part a at all for me. $\endgroup$ – hmtkd Feb 12 at 3:16

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