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I am trying to find the conditional PMF of a multinomial analytically and though I know my result is wrong I can't seem to pinpoint where my argument is wrong. Seeking help to find my mistake.

Given: $\vec{X} = Mult(n,\vec{p})$ where $\vec{X} = [X_1,X_2,X_3,...X_k]$ and $\vec{p} = [p_1,p_2,p_3,...p_k]$. Find the conditional PMF of $\vec{X}$ given $X_1 = n_1$

I have tried to solve it like this:

A bag contains $n$ different balls belonging to $k$ categories and you know the PMF of the number of balls belonging to each category. A third person goes through the bag and removes all balls belonging to type 1 ($n_1$ in count) and you have to find the PMF of balls that remain in the bag

$E_1$ = A third person goes through the bag and finds $n_1$ balls belonging to $X_1$ and removes them

$E_2$ = You choose a random ball from the bag

To find: $P(E_2 \in type 2|E_1)$

Solution: Using baye's theorem:

$$P(E_2 \in type 2|E_1) = \frac{P(E_1|E_2 \in type 2)*P(E_2 \in type 2)}{P(E_1)}$$ $$ = \frac{{n-1 \choose n_1}(p_1)^{n_1}(1-p_1)^{n-1-n_1}*{p_2}}{{n \choose n_1}(p_1)^{n_1}(1-p_1)^{n-n_1}}$$ $$\frac{(n-n_1)*p_2}{n*(1-p_1)}$$

Which is clearly not right. It cannot depend on $n_1$. Can someone please point me to what i am going wrong?

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The joint probability mass function of the vector $\mathbf X$ is $$p(x_1,\ldots,x_k)=\binom{n}{x_1\,\cdots\,x_k}p_1^{x_1}\cdots p_k^{x_k}\mathbb I_{x_1+\ldots+x_k=n}$$ Hence $$\mathbb P((X_1,\ldots,X_k)=(x_1,\ldots,x_k))=\binom{n}{x_1\,\cdots\,x_k}p_1^{x_1}\cdots p_k^{x_k}$$ if $n_1+\ldots+x_k=n$ and the probability is zero otherwise. This implies that \begin{align*}\mathbb P((X_1,\ldots,X_k)=(x_1,\ldots,x_k)\vert X_1=x_1) &\propto \mathbb P((X_1,\ldots,X_k)=(x_1,\ldots,x_k)\\ &\propto \dfrac{n!}{x_1!\cdots x_k!}p_1^{x_1}\cdots p_k^{x_k}\\ &\propto \dfrac{1}{x_2!\cdots x_k!}p_2^{x_2}\cdots p_k^{x_k}\\ &=\dfrac{(n-x_1)!}{x_2!\cdots x_k!}[p_2/\bar p_1]^{x_2}\cdots [p_k/1-p_1]^{x_k} \end{align*} the normalisation emerging from the Multinomial theorem.

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