#### robespierre722

##### New member

- Joined
- Mar 17, 2019

- Messages
- 4

3 people(A,B,C) enter an elevator at the ground floor of a 5-storey building, another 3 people(D,E,F) enter in another elevator.

Each person is independent of the others to get off at any of the 5 floors above ground.

a)What is the probability that exactly 2 people will get off at the 3rd floor ? (regardless of the elevator)

b)What is the probability that no person from the first elevator will want to go at the 5th floor.

a) So there are 5 possibilities for each person -> 5^6 possible cases

There are 6C2=15 possible pairs for that 3rd floor .

And now the funny part, i was presuming that for example on the 1st floor, the 4 people left are going to be able to enter in (4C1+4C2+4C3+4C4)=15 ways since they can enter 1 or 2 or 3 or 4 people . But there are 4 floors ( i am not counting the 3rd floor) -> (4C1+4C2+4C3+4C4)*4=60

And my answer is ((4C1+4C2+4C3+4C4)*4*6C2)/(5^6)=0.0576 which is the wrong answer . If i put (4P1+4P2+4P3+4P4)*5, P instead of C and 4 instead of 5 -> the good answer but i don't understand why .

b)5^6 possible cases. From the first elevator 4^3 since they can't enter 5th floor and from the second elevator 5^3 -> (5^3)*(4^3)/5^6 but it is the wrong answer again .