Combinatorial Probability Question Similar to Sheldon Ross's A First Course in Probability 10th ed Chapter 2 Problem 7

Question

I wrote a combinatorial probability question, and I am not sure if my solution is correct. It is just a slight rewording of a problem from Sheldon Ross's A First Course in Probability 10th ed. Chapter 2 Problem 7. I would like to know your thoughts.

Question:

"Vince's carnival choo-choo train containing 16 people first stops at station 1 and stops last at station 5. After station 5, it must be completely empty. People can leave at any station but not board, and each person is equally likely to leave at any stop. Additionally, the passengers are leaving independent of each other. What is the probability that exactly half of the people have left by station 2?"

My Solution

First, we calculate the number of ways the event could take place.

Let $x_i$ be the amount of people who left at station $i$, where $i \in\{1,2,3,4,5\}$.

Firstly, there are ${{16}\choose{8}}$ ways to choose the eight people that will have left after station two. We then consider a specific $(x_1,x_2)$. From the 8, there are ${{8}\choose{x_1}}$ for people to leave at station 1, and there are ${{8-x_1}\choose{x_2}}$ at station 2. Then, we sum through all possible $x_1$'s and $x_2$'s that satisfy $x_1+x_2=8$. Giving us:

$$\sum_{x_1+x_2=8}{{8}\choose{x_1}}{{8-x_1}\choose{x_2}}$$ Multiplying them together, we have: $$ {{16}\choose{8}}\sum_{x_1+x_2=8}{{8}\choose{x_1}}{{8-x_1}\choose{x_2}}={{16}\choose{8}}\sum_{x_1+x_2=8}\frac{8!}{{x_1}!{x_2}!}={{16}\choose{8}}2^8 $$

Next, we compute the ways for the remaining 8 people to leave the last three stations. For some $(x_3,x_4,x_5)$, there are ${{8}\choose{x_3}}{{8-x_3}\choose{x_4}}{{8-x_3-x_4}\choose{x_5}}$ ways for the remaining people to leave within stations 3 to 5. Summing through all possible $x_3,x_4,$ and $x_5$ such that $x_3+x_4+x_5=8$, we have:

$$\sum_{x_3+x_4+x_5=8}{{8}\choose{x_3}}{{8-x_3}\choose{x_4}}{{8-x_3-x_4}\choose{x_5}}=\sum_{x_3+x_4+x_5=8}\frac{8!}{{x_3}!{x_4}!{x_5}!}=3^8$$

There are ${{16}\choose{8}}{2^8}{3^8}$ outcomes in the event.

Finally, we calculate the number of possible outcomes in the entire sample space.

Following the same argument as above, given a specific $(x_1,x_2,x_3,x_4,x_5)$ there are ${{16}\choose{x_1}}{{16-x_1}\choose{x_2}}{{16-x_1-x_2}\choose{x_3}}{{16-x_1-x_2-x_3}\choose{x_4}}{{16-x_1-x_2-x_3-x_4}\choose{x_5}}$ ways for the 16 people to leave within stations 1 to 5.

Summing through all possible $x_1,x_2,x_3,x_4,$ and $x_5$ such that $x_1+x_2+x_3+x_4+x_5=16$, we have:

$$\sum_{x_1+x_2+x_3+x_4+x_5=16}{{16}\choose{x_1}}{{16-x_1}\choose{x_2}}{{16-x_1-x_2}\choose{x_3}}{{16-x_1-x_2-x_3}\choose{x_4}}{{16-x_1-x_2-x_3-x_4}\choose{x_5}}$$

$$=\sum_{x_1+x_2+x_3+x_4+x_5=16}\frac{16!}{{x_1}!{x_2}!{x_3}!{x_4}!{x_5}!}=5^{16}$$

So the probability is:

$$\frac{{{16}\choose{8}}{2^8}{3^8}}{5^{16}}\approx 0.1417$$

Edit: I added necessary assumptions based on the comments

Answer 1

To summarize the discussion in the comments:

the problem needs clarification. First of all, it seems we are to assume a uniform distribution and independence, which is far from obvious (in real life, people travel in groups and, of course, some stops will be more popular than others).

Secondly, I would read the problem as requiring that at least $8$ people leave at one of the first two stations, not exactly $8$.

In any case:

Assuming uniformity and independence, there is a $\frac 25$ chance that a given passenger gets off at one of the first two stops. It follows that the probability that at least $8$ people do that is:

$$\sum_{k=8}^{16}\binom {16}k \times \left(\frac 25\right)^k\times \left(\frac 35\right)^{16-k}$$

If we want exactly $8$, then we just take the term corresponding to $k=8$, $$\binom {16}8 \times \left(\frac 25\right)^8\times \left(\frac 35\right)^{8}$$

which matches your result.