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]]