Given 7 experiments and 3 types of successes, with probabilities $p_1, p_2, p_3$ respectively and $p_1+ p_2+ p_3 = 1$ we'll denote:
$$X_i = \text{number of successes of type i}$$
I'm looking for $P(X_1 = 3 | X_3 \le 2)$
So by definition of conditional probability:
$$P(X_1 = 3 | X_3 \le 2) = \frac{P(X_1 = 3, X_3 \le 2)}{P(X_3 \le 2)} = $$
$$\frac{P(X_1 = 3, X_2 = 2, X_3 = 2) + P(X_1 = 3, X_2 = 3, X_3 = 1) + P(X_1 = 3, X_2 = 4, X_3 = 0)}{P(X_3 = 2) + P(X_3 = 1) + P(X_3 = 0)}$$
We know that $(X_1, X_2, X_3) \sim{Multi(7, p_1, p_2, p_3)}$
And $X_i \sim{Bin(7, p_i)}$
So from here we just use the formulas.
My question is if this could be simplified, or is there an immediate formula for $(X_i | X_j \le x_j)$ ?