Exercise 1.23

Question

Three people get into an empty elevator at the first floor of a building that has 10
floors. Each presses the button for their desired floor (unless one of the others has
already pressed that button). Assume that they are equally likely to want to go to
floors 2 through 10 (independently of each other). What is the probability that the
buttons for 3 consecutive floors are pressed?

fifthist x · Accepted Answer

We are interested in the case of $3$ consecutive floors. There are $7$ equally 
likely possibilities $$(2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, 7), 
(6, 7, 8), (7, 8, 9), (8, 9, 10).$$

For each of this possibilities, there are $3$ ways for $1$ person to choose 
button, $2$ for second and $1$ for third ($3!$ in total by multiplication rule).

So number of favorable combinations is $$7 * 3!$$

Generally each person have $9$ floors to choose from so for $3$ people there are 
$9^3$ combinations by multiplication rule.

Hence, the probability that the buttons for $3$ consecutive floors are pressed
is $$\frac{7 * 3!}{9^3}$$

Exercise 1.23

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Exercise 1.23

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