Intro:
Thanks for reading.
My friend and I were working on a probability question. We got different answers. However, I'm still a beginner in probability, and can't see which way is correct, which way is incorrect, and why…
Question:
There are 3 people in an elevator that will stop at 3 floors. The people get off at the floors randomly. They're all getting off somewhere (the elevator will be empty at the end).
What's the probability that exactly one person gets off at each floor?
Things we both did:
We both computed the probability of exactly one person getting off at the first floor, $P_1$ , one person getting off at the second floor given that one person got off at the first floor, $(P_2 | \mathrm{1PersFloor1})$, and the probability of one person getting off at the third floor given that one got off at the first and one got off at the second, $(P_3 | (\mathrm{1PersFloor1\ \cap \ 1PersFloor2}))$.
We then multiplied them all together to get the total probability.
My Attempt:
At the first floor, the probability that any given person gets off, and the other two don't, would be $\frac{1}{3}*(\frac{2}{3})^2$, since the first person chooses the first floor with probability $(\frac{1}{3})$ and the other two don't choose the first floor with probability $(\frac{2}{3})$.
Additionally, there are $\binom{3}{1}$ ways to choose the person that gets off at the first floor. So the total probability of exactly one person getting off on the first floor would be:
$$P_1=\binom{3}{1}*\frac{1}{3}*(\frac{2}{3})^2 = \frac{4}{9}$$
By the same logic, on the second floor…
$$(P_2 | \mathrm{1PersFloor1})=\binom{2}{1}*\frac{1}{2}*(\frac{1}{2}) = \frac{1}{2}$$
And by the same logic, on the third floor…
$$(P_3 | (\mathrm{1PersFloor1\ \cap \ 1PersFloor2}))=\binom{1}{1}*1=1$$
Finally, we multiply them all together and get:
$$\frac{4}{9} * \frac{1}{2}*1=\frac{2}{9}$$
My Friend's Way:
On the first floor, either one person can get off, two people can get off, or three people can get off.
The probability that only one person gets off is the number of total ways one person can get off, divided by the total number of things that can happen!
There are $\binom{3}{1}$ ways to choose the single person that gets off.
There are $\binom{3}{1} + \binom{3}{2} + \binom{3}{3}$ total things that can happen, since either one person can get off, two people can get off, or three people can get off.
Therefore, the probability that exactly one person gets off at the first floor is:
$$P_1=\frac{\binom{3}{1}}{\binom{3}{1} + \binom{3}{2} + \binom{3}{3}}=\frac{3}{7}$$
Applying the same logic to the second floor, we get…
$$(P_2 | \mathrm{1PersFloor1})=\frac{\binom{2}{1}}{\binom{2}{1}+\binom{2}{2}}=\frac{2}{3}$$
And to the third floor…
$$(P_3 | (\mathrm{1PersFloor1\ \cap \ 1PersFloor2}))=\frac{\binom{1}{1}}{\binom{1}{1}}=1$$
Multiplying all three probabilities together, we get:
$$\frac{3}{7}*\frac{2}{3}*1=\frac{6}{21}$$
My Thoughts:
I think the way I did it is correct…actually, I looked at the book and it agrees with me.
However, I can't see why the way he did it is wrong…
I feel like it has to do something with the underlying assumptions that I use and that he uses being different…but I'm not sure how to explain why to him (or to myself, for that matter).
I would love to know both why the correct way is the correct way, and why the incorrect way isn't.
Final Notes:
I tagged this as "Problem-Solving" because I really want to understand the underlying assumptions in both methods, why one works and one doesn't – not just the answer to the question.
Thanks again!
Best Answer
Your friend is assuming that the various combinations of people that can get off on the first floor are equally probable, which is not correct. The chance that all three get off is $\left(\frac13\right)^3=\frac 1{27}$, while the chance that a particular person is the only one to get off is $\frac 13\cdot \left(\frac23\right)^2=\frac 4{27} $
Your friend also neglected the possibility that nobody gets off on the first floor.
I think an easier approach is to note that there are $3^3$ ways that people can get off and $3!=6$ of them have one per floor, so the chance is $\frac 6{27}=\frac 29$