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 $1\dfrac{2}{3}\ min$

Step-by-step 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 = $\dfrac{2}{3}$

Probability that a person choose the slow elevator = $\dfrac{1}{3}$

So, the expected waiting time would be

$E[x]=\sum xp(x)=3\times \dfrac{1}{3}+1\times \dfrac{2}{3}\\\\=1+\dfrac{2}{3}\\\\=\dfrac{3+2}{3}\\\\=\dfrac{5}{3}\\\\=1\dfrac{2}{3}\ min$

Hence, the expected waiting time is $1\dfrac{2}{3}\ min$