# Probability that no 2 people get off at some stop

• Oct 29th 2009, 09:15 AM
ghostb
Probability that no 2 people get off at some stop
There is a bus with 6 people, and the bus will stop 10 times. All passengers are equally likely to get off anywhere. What is the prob. that no 2 people get off at the same stop?

I have C(6 10)= 10factorial/(6factorial*4factorial)=210 ways 6 people can get off at 10 stops.

But from there, I am stuck. I'm unsure of how to find whether 2 people will get off at the same stop. Any assistance is greatly appreciated.

Thanks.
• Oct 29th 2009, 02:01 PM
Soroban
Hello, ghostb!

Quote:

There is a bus with 6 people, and the bus will stop 10 times.
All passengers are equally likely to get off anywhere.
What is the probability that no two people get off at the same stop?

The 1st person has 10 choices of stops: .$\displaystyle \frac{10}{10}$

The 2nd person has 9 choices of stops: .$\displaystyle \frac{9}{10}$

The 3rd person has 8 choices of stops: .$\displaystyle \frac{8}{10}$

The 4th person has 7 choices of stops: .$\displaystyle \frac{7}{10}$

The 5th person has 6 choices of stops: .$\displaystyle \frac{6}{10}$

The 6th person has 5 choices of stops: .$\displaystyle \frac{5}{10}$

Therefore: .$\displaystyle P(\text{6 different stops}) \;=\;\frac{10}{10}\cdot\frac{9}{10}\cdot\frac{8}{1 0}\cdot\frac{7}{10}\cdot\frac{6}{10}\cdot\frac{5}{ 10} \;=\;0.1512$

• Oct 29th 2009, 02:46 PM
ghostb
Hmmm.... does that give the probability that no 2 passengers get off at the same stop? That seems way too easy.
• Oct 29th 2009, 03:16 PM
Plato
Quote:

Originally Posted by ghostb
Hmmm.... does that give the probability that no 2 passengers get off at the same stop? That seems way too easy.

Just to be sure that we are asking the same question, let's write it in English.
That is the probability that no two people exit at the same stop.

Now I have seen this same setup, but the question was:
What is the probability that two people exit at the same stop?