Question

A building contains two elevators, one fast and one slow. The average waiting time for the slow elevator is 3 min. and the average waiting time of the fast elevator is 1 min. If a passenger chooses the first elevator with probability 2 3 and the slow elevator with probability 1 3 , what is the expected waiting time?

Answer 1

Answer: The expected waiting time is
`1(2)/(3)\ min`

Explanation:

Since we have given that

Average waiting time for slow elevator = 3 min

Average waiting time for fast elevator = 1 min

probability that a person choose the fast elevator =
`(2)/(3)`

Probability that a person choose the slow elevator =
`(1)/(3)`

So, the expected waiting time would be

`E[x]=\sum xp(x)=3* (1)/(3)+1* (2)/(3)\\\\=1+(2)/(3)\\\\=(3+2)/(3)\\\\=(5)/(3)\\\\=1(2)/(3)\ min`

Hence, the expected waiting time is
`1(2)/(3)\ min`