suppose we throw n balls( note we will throw exactly n balls, and each throwing

Question

suppose we throw n balls( note we will throw exactly n balls, and each throwing is independant of other throwing) into n bins with the probability of a ball landing in each of the n bins being equal.  We assume each throwing is independant of the other throwing.  You can assume n is large.  You may need the following mathematical fact: when n -> infinity, (1-1/n)^n -> e^-1, where e is the well-known mathematical constant.  What is the probability of a particular box ( say the first box) end up being empty after the n throwing?  What is the expected number of empty bins?

Explanation / Answer

Since there are n number of bins and the probability of a ball going into any bin is equal

so the probability of a ball landing in any bin is 1/ n

So in the first throw, probability of the first bin being empty is (1 - 1/n)

in the 2nd throw probability of the first bin being empty = (1 - 1/n)

and so on till the nth throw, probability = (1 - 1/n)

hence the combined probbaility = product of all the cases we derived above

= (1 - 1/n) ^ n

This is the combined probability that the 1st bin will remain empty after n throws

Now if n tends to infinity. according to the given formula, (1 - 1/n)^n = e^-1

hence the required probability is e^-1 = 0.3679

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