Exactly one of six similar keys opens a certain door. If you try the keys, one after another, what is the expected number of keys that you will have to try before success?

References

Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.

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]]