Question

An elevator has 4 passengers and 8 floors. Find the probability that no 2 passengers get off on the same floor considering that it is equally likely that a person will get off at any floor.

Answer 1

The probability is 5/343.

Given :8 floors and 4 passengers.

Passengers will be assumed to start on the first floor, where everyone goes up, and each passenger only gets off on the 2nd floor (no one gets off on the 1st floor!)

Now we need to find the number of all ordered lists of four numbers out of seven, eg 2<3<7<8, 3<6<7<8, 4<5<6o<8, etc.

This can be done in 7C4 ways.

The total numbers of ways 4 passengers can get off at 7 floors (starting at the 2nd!) is $7^{4}$.

Therefore the probability is $\frac{7C4}{7^{4} }$ = $\frac{5}{343}$

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