Probability of 9 persons walking into a 3-carriage train. Nine persons go into a 3-carriage tram. Each person chooses the carriage at
random. What is the probability that there will be 3 persons in each carriage?
The probability of a person entering any carriage is $ \frac{1}{3} $. I thought of solving it using the multinomial scheme, which gave me $$\frac{9!}{3!3!3!} \Bigl(\frac{1}{3}\Bigr)^9=35 \cdot\frac{2^4}{3^7}$$
However the book lists $\frac{25}{4}\cdot\bigl(\frac{2}{3}\bigr)^7$ as the answer. Where do i go wrong?
 A: I agree with you.  There are $3^9$ ways to arrange the people.  The number of ways to get $3-3-3$ is ${9 \choose 3}$ for the first car times ${6 \choose 3}$ for the second.
A: You are right (modulo the fact that $\frac{9!}{3!3!3!}\left(\frac{1}{3}\right)^{9}\neq 35\cdot \frac{2^4}{3^7}$, but $35\cdot\frac{2^4}{3^8}$). Indeed, the probability of event $E_1$ — that carriage #1 has exactly 3 people inside is 
\begin{align*}
\mathbb{P}\left[E_1\right] &= \binom{9}{3}\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^6
\end{align*}
Conditioned on $E_1$, the probability of $E_2$ — that carriage #2 has exactly 3 people inside becomes
\begin{align*}
\mathbb{P}\left[E_2 \mid E_1\right] &= \binom{6}{3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^3
\end{align*}
Conditioned on $E_1\cap E_2$, the probability of $E_3$ — that  that carriage #3 has exactly 3 people inside is just 
\begin{align*}
\mathbb{P}\left[E_3 \mid E_1\cap E_2\right] &= 1
\end{align*}
The probability you're looking for is obtained by multiplying the 3 (by definition of conditional probability):
\begin{align*}
\mathbb{P}\left[E_1\cap E_2 \cap E_3\right] &= \mathbb{P}\left[E_3 \mid E_1\cap E_2\right]\cdot\mathbb{P}\left[E_2 \mid E_1\right]\cdot\mathbb{P}\left[E_1\right] \\
&=
\binom{9}{3}\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^6\cdot \binom{6}{3}\left(\frac{1}{2}\right)^6 \cdot 1 = \binom{9}{3}\binom{6}{3}\left(\frac{1}{3}\right)^{9} \\
&=\frac{9!}{3!3!3!}\left(\frac{1}{3}\right)^{9}
\end{align*}
