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| rev | Admin | (Created page with "'''Solution: C''' The probability that you need to try <math>k </math> keys equals <math display = "block"> \frac{5}{6} \frac{4}{5} \cdots \frac{6-k-1}{6-k} \frac{1}{6-k-1} = \frac{1}{6}. </math> Hence the probability distribution is uniform on <math>\{1,\ldots,6\}</math>. Then the expected value equals <math display = "block"> \frac{1}{6}\sum_{k=1}^6 k = 3.5. </math>") | Jun 26'24 at 1:36 | +377 |