Probability and statistics

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Quote:

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 . . .

Quote:

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%.

Quote:

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