Mr. Smith works on the 13th floor of a 15 floor building. The only elevator moves continuously through floors 1, 2, . . . , 15, 14, . . . , 2, 1, 2, . . . , except that it stops on a floor on which the button has been pressed. Assume that time spent loading and unloading passengers is very small compared to the travelling time.
Mr. Smith complains that at 5 pm, when he wants to go home, the elevator almost always goes up when it stops on his floor. What is the explanation?
Now assume that the building has n elevators, which move independently. Compute the proportion of time the first elevator on Mr. Smith’s floor moves up.

Step-by-step explanation:

Solution:

Think of it this way. The elevator could be on

floor 1 moving up

floor 2 moving up

floor 2 moving down

floor 3 moving up

floor 3 moving down

floor 4 moving up

floor 4 moving down

5 up

5 down

6 up  

6 down

7 up  

7 down

8 up

8 down

9 up

9 down

10 up

10 down

11 up

11 down

12 up

12 down

- All of these would mean that the guy on the 13th floor would get on an elevator that was moving up.

The rest :-

13 moving up

13 moving down

14 moving up

14 moving down

15 moving down

- 13 up is still moving up but the other 4 possibilities is the only way for the elevator to be coming down when he gets on.

- So there are 28 possibilities and only 4 where the guy is lucky. So the probability that he gets an elevator coming down is 4/28 or 1/7leaving 6/7 that the elevator is moving up.

- As for the second part of question, Since the elevators are independent and you are asking only about one of the elevators, each one of those elevators moves up 6/7 of the time, regardless of how many elevators.

- Now if the question said that you had 7 elevators stop on the floor all at the same time, the probability would say that 1 of the seven would be going down.

step-by-step explanation: