Conditional probability in multinomial distribution Consider a multinomial distribution with $r$ different outcomes, where the $i$th outcome having the probability $p_i$, $i$=1,...,$r$, $\sum_{i=1}^r p_i = 1$. Denote $X_i$ be the number of times the $i$th type outcome occurs, $i$=1,...,r.
I find it difficult to construct the conditional probability $P(X_i=x_i, i=1,...,r-1\mid X_r=j)$, for $0 \le j \le n$.
Here is my work:
I first use the definition of conditional probability.
$$P(X_i=x_i\mid X_r=j)= \frac{P(X_i=x_i \cap X_r=j)}{P(X_r=j)}$$
Now, for the numerator, I use the multinomial distribution, which gives
$$P(X_i=x_i \cap X_r=j) = \frac{n!}{x_i! j!} p_i^{x_i} p_r^j$$
For the denominator, I write
$$P(X_r=j) =\frac{n!}{j! (n-j)!} p_r^j (1-p_r)^{n-j}$$
Combining the above, I then conclude
$$P(X_i=x_i\mid X_r=j)= \frac{(n-j)! p_i^{x_i}}{x_i! (1-p_r)^{n-j}}$$
I am completely new in learning probability theory and I welcome any comments and ideas towards my work and the clarity of the question. Thank you.
 A: 
Now, for the numerator, I use the multinomial distribution, which gives
  $$P(X_i=x_i \cap X_r=j) = \frac{n!}{x_i! j!} p_i^{x_i} p_r^j$$

Actually, using the multinomial distribution, one gets
$$P(X_i=x_i \cap X_r=j) = \frac{n!}{x_i! j!\color{red}{(n-x_i-j)!}} p_i^{x_i} p_r^j\color{red}{(1-p_i-p_r)^{n-x_i-j}}.$$
Then one should get that the conditional distribution of $X_i$ conditionally on $[X_r=j]$ is binomial $(k,q)$, for some well chosen parameters $k$ and $q$.
Note also that you mention the joint conditional distribution of $(X_i)_{i\ne r}$ conditionally on $[X_r=j]$. True, this can be computed following the same idea, but this question is not addressed simply by computing $P(X_i=x_i \cap X_r=j)$.
A: Under condition $X_{r}=j$ the distribution of $\left(X_{1},\dots,X_{r-1}\right)$
is again multinomial. This with parameters $n-j$, $q_{1},\dots,q_{r-1}$
where $q_{i}=\frac{p_{i}}{1-p_{r}}$.
This leads to $$P\left\{ X_{1}=x_{1},\dots,X_{r-1}=x_{r-1}\mid X_{r}=j\right\} =\frac{\left(n-j\right)!}{x_{1}!\cdots x_{r-1}!}q_{1}^{x_{1}}\cdots q_{r-1}^{x_{r-1}}$$
for nonnegative integers $x_{1},\dots,x_{r-1}$ that satisfy $x_{1}+\cdots+x_{r-1}=n-j$.

It comes to working out:
$P\left\{ X_{1}=x_{1},\dots,X_{r-1}=x_{r-1}\mid X_{r}=j\right\} =\frac{P\left\{ X_{1}=x_{1},\dots,X_{r-1}=x_{r-1}\wedge X_{r}=j\right\} }{P\left\{ X_{r}=j\right\} }=\left[\frac{n!}{x_{1}!\cdots x_{r-1}!j!}p_{1}^{x_{1}}\cdots p_{r-1}^{x_{r-1}}p_{r}^{j}\right]/\binom{n}{j}\left(1-p_{r}\right)^{n-j}p_{r}^{j}$
