Ten passengers randomly enter three train cars. What is the probability that the first car will contain three passengers?
I approached this problem following a line of reasoning from David Morin's book Probability: For the Enthusiastic Beginner:
We have 10 passengers and 4 boundaries, so a total of 14 spots need to be filled. However, the first and last spots are for the beginning boundary of the first car and the last boundary of the last car respectively. Furthermore 3 passengers will be in the first car, so the situation looks as follows:
| X X X | _ _ _ _ _ _ _ _ |
Then, the problem is in how many was can that last boundary be placed to satisfy the problem conditions. There are 8 spot lefts for 1 boundary, so this should be $\binom81$.
For the total number of possible combinations, the problem looks as follows:
| _ _ _ _ _ _ _ _ _ _ _ _ |
Now there are 12 possible spots in which to put the two internal car boundaries, so $\binom{12}2$.
The probability then of having 3 passengers in the first car is given by: $$P=\frac{\binom81}{\binom{12}2}=0.12$$ However, the professor gave the following answer:
All cars are equally likely so that the first car is selected with probability $p=\frac13$. This gives the desired probability to be: $$\binom{10}3*\left(\frac13\right)^3\left(\frac23\right)^7=0.26$$ I am interpreting that answer as: let $P(k)$ be the probability of having $k=3$ passengers in the first car (with the probability of a passenger being in the first car as $\frac13$ and the probability of a passenger being in any of the other cars being $\frac23$). Then the experiment is a Bernoulli repeated trial experiment, and using the above result is given by the appropriate formula, $$\binom nkp^kq^{n-k}$$ My question is, where did my original interpretation go wrong, based on the problem statement?
You way only works if the passengers are indistinguishable. Since people arent indistinguishable, you can't use the stars and bars method that you did.
The reason there is a $\binom{10}{3}$ on that Bernoulli formula, is because you have to account for all the different ways in which 3 people can end up in the selected car, because people are all different.