Exercise


ABy Admin
Jun 26'24

Answer

Solution: C

The probability that you need to try [math]k [/math] keys equals

[[math]] \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]] \frac{1}{6}\sum_{k=1}^6 k = 3.5. [[/math]]

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