1. ## Probability and statistics

Ten people get on an elevator on the first floor of a six story building. Assume that each of the 10 people get off at random at any one of the five upper floors.

a) What is the probability that three or more people get off at least one floor?
b) Assume that the ten people are actually four couples and two of the couples have one child each. Assume also that families get off at the same floor. What is the probability that three or more people get off at least one or more floors?
c) Again assume the ten people are four couples and two children. What is the probability that the two children get off on different floors?

2. Hello, wvlilgurl!

Ten people get on an elevator on the first floor of a six-story building.
Assume that each of the 10 people get off at random at any one of the five upper floors.

a) What is the probability that 3 or more people get off at least one floor?
Each of the 10 people have a choice of 5 floors.
. . There are $5^{10}$ possible outcomes.

The opposite of "three or more people" is "two or less people".

The only way that two or less people get off on all the floors
. . is if two people get off on each floor.

This can happen in ${10\choose2,2,2,2,2} \:=\: 113,400$ ways.

Hence: . $P(\text{2 or less}) \;=\;\frac{113.400}{5^{10}}$

Therefore: . $P(\text{3 or more}) \;=\;1 - \frac{113,400}{5^{10}} \;=\;\frac{9,655,225}{9,765,625}$

I'll let you reduce the fraction . . .

b) Assume that the 10 people are actually 4 couples and 2 of the couples have 1 child each.
Assume also that families get off at the same floor.
What is the probability that 3 or more people get off at least one or more floors?
The families are: . $\{A,B,c\},\;\{D,E,f\},\;\{G,H\},\;\{I,J\}$

A family-of-three will get off on one the floors, right?

The probability is: 1.00 = 100%.

c) Again assume the 10 people are 4 couples and 2 children.
What is the probability that the 2 children get off on different floors?
There are four families: . $\{ABc\},\;\{D{E}f\},\;\{GH\},\;\{IJ\}$
. . They can get off in $5^4 \:=\:625$ ways.

Suppose the two children get off on the same floor.
Then we have three "families": . $\boxed{\{ABc\}\{D{E}f\}},\;\{GH\},\;\{IJ\}$
. . They can get off in: $5^3 = 125$ ways.

Hence, the two children get off on different floors in: $625 - 125 \:=\:500$ ways.

Therefore: . $P(\text{children, different floors}) \;=\;\frac{500}{625} \;=\;\frac{4}{5}$