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I'm struggling to understand how to solve the following problem.

We have (1,2,3,4) is a multinomial distributed r.vs with parameters =10 and p =(1 =0.3,2=0.3,3=0.3,4=0.1).

Now I need to find P(1=3,2=3,3=3|4=1). I would know how to approach this if we had for example P(1=3|4=1) using: enter image description here

And then just treating it as a binomial distribution, but what to do if we need to find the probability of more than 1 variable occurring, given that 1 other has occurred?

LushyIvy
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1 Answers1

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The formula you included tells you that the conditional distribution of the other $X_i$, given $X_4$, is also multinomial, not just binomial. In other words, $$((X_1, X_2, X_3)^\top \mid X_4 = 1) \sim \operatorname{Multinomial}(n = 9; p_1 = p_2 = p_3 = 1/3).$$ So the desired probability is simply $$\Pr[X_1 = X_2 = X_3 = 3 \mid X_4 = 1] = \frac{9!}{3!3!3!} (1/3)^3 (1/3)^3 (1/3)^3.$$

heropup
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