Question 570976: Suppose k identical boxes contained n balls numbered one through n. One ball is drawn from
each box. What is the probability that m is the largest number drawn?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! For this problem we will assume that  .
The best way is probably to try small cases (for m). Let P(m) be the probability that m is the largest number drawn. If m = 1, then every number drawn must be a 1, and
 = (\frac{1}{n})^k)
If m = 2, every number must be a 1 or a 2, and we would try to claim something like
 = (\frac{2}{n})^k)
However, this is incorrect, because this includes the case where every number is a 1 (which violates the constraint that at least one number must be a 2). To fix this, we subtract P(1):
 = (\frac{2}{n})^k - P(1) = (\frac{2}{n})^k - (\frac{1}{n})^k)
Finding P(3) is similar to finding P(2), except that we subtract P(2):
 = (\frac{3}{n})^k - P(2) = (\frac{3}{n})^k - (\frac{2}{n})^k +(\frac{1}{n})^k)
This generalizes to
 = (\frac{m}{n})^k - P(m-1)) , or
 = (\frac{m}{n})^k - (\frac{m-1}{n})^k + (\frac{m-2}{n})^k ... \pm (\frac{1}{n})^k)
where the plus/minus of (1/n)^k depends on the parity of m (i.e. if m is odd, +; if m is even, -).
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