Probability Question. Which solution is right? Suppose, there is a building with $6$ floors and a ground floor. If $10$ persons get into an elevator on the ground floor, what is the probability that exactly $2$ persons will get out on the $2nd$ floor?
I can present two solutions, but which one of them is right?
Let $x_i$ denote the number of persons getting out on the $ith$ floor. Then $x_1+x_2+x_3+x_4+x_5+x_6=10$ has ${15 \choose 5}$ non-negative solutions. If $x_2=2$, then there are ${12 \choose 4}$ favourable solutions. Hence, required probability = $\frac{{12 \choose 4}}{{15 \choose 5}}$
Another solution can be,
Total number of ways in which $10$ people can get out is $6^{10}$.
Number of ways in which exactly two people can get out on 2nd floor is ${10 \choose 2}*5^8$.
Hence, required probability is $\frac{{10 \choose 2}*5^8}{6^{10}}$.
Also, the two solutions don't match after calculation.
 A: Your second approach is correct.  
The events in your first approach are not equally likely to occur.  There is only one way for all ten people to exit the elevator on the fifth floor.  However, there are $\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{1}\binom{1}{1}$ for two people each to exit on the third, fourth, fifth, and sixth floors and for one person each to exit on the first and second floors.
A: The first solution is incorrect, because it implies that each person’s decision to stay or get out of the elevator is not independent from other persons’ decision. Think about it like this, You have $n$ identical coins. You cannot differentiate each from the others. But still, since the result of individual coin flip is independent, we are less likely to get all heads or all tails than $2$ heads and $n-2$ tails for example.
The second solution is correct. However, I want to share something I am tinkering with.
The number of ways $x_{i}$ people get out on $i$-th floor is the coefficient of $a_{1}^{x_{i}}a_{2}^{x_{2}}a_{3}^{x_{3}}a_{4}^{x_{4}}a_{5}^{x_{5}}a_{6}^{x_{6}}$ from $(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6})^{10}$. Since we are only interested in $x_{2}=2$, we just need to calculate coefficient of $a_{2}^{2}$ from $(1+a_{2}+1+1+1+1)^{10}=(a_{2}+5)^{10}$, which is $\binom{10}{2}5^{8}$.
