Revision as of 00:36, 26 June 2024 by 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>")
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]]