About ten percent of users do not close Windows properly. Suppose that Windows is installed in a public library that is used by random people in a random order. (a) On the average, how many users of this computer do not close Windows properly before someone does close it properly? (

Let X be the number of users that do not close Windows properly before someone does including that

person. Then X has a geometric distribution with a success rate of p = 0.9. The expected value of

X is given as 1/p = 1.1111. Define a random variable Y to be the number of users that do not close

windows properly before someone does, not including the final person who closes windows properly. If

X = 1, then Y = 0; if X = 2, then Y = 1; if X = 3, then Y = 2, etc. In general, if X = k, then

Y = k − 1, so we can write Y as a function of X: Y = X − 1. Therefore, E[Y ] = E[X] − 1 = .11111.

The random variable Y has a modified geometric distribution with p = .9.

Explanation:

i do not get the question.