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I am studying for Exam P to take in a few hours so I'd appreciate very much if someone got to me befoe I left. The problem says that we have three categories of risk, high, medium and low risk. The portion of people that fall into the high risk category is $20%$, medium is $40%$, and low is also $40%$. A random sample of four policyholders is chosen. What is the probability that the number of low risk policy holders exceeds the number of medium risk policyholders by at least 2? The solutions say this is trinomial problem where $$P(H=h,M=m,L=l)= {n \choose h}{{n-h} \choose m}(p_H)^h(p_M)^m(p_L)^l$$ I don't see where this distribution is come up with. It doesn't make sense to me. I feel it should be: $$P(H=h,M=m,L=l)= {n \choose h}{{n-h} \choose m}{{n-h-m} \choose l}(p_H)^h(p_M)^m(p_L)^l$$ Is there any way to explain this?

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up vote 1 down vote accepted

Since $n=h+m+l$, we have $n-h-m=l$.

Thus your term $\binom{n-h-m}{l}$ is just $\binom{l}{l}$, which is $1$. So your formula and the given formula give exactly the same answer.

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so really we could rearrange it as well to say $$P(H=h,M=m,L=l)= {n \choose l}{{n-l} \choose h}(p_H)^h(p_M)^m(p_L)^l$$ Would these be the same? If we wanted to go that route? – TheHopefulActuary Jan 17 '13 at 16:11
Sure, you can permute the $h$, $l$, $m$ in any way you wish. In all cases the expression "simplifies" to $\frac{n!}{h!l!m!}$. – André Nicolas Jan 17 '13 at 16:14

By analogy, a binomial distribution looks like

$$\binom{n}{h} (p_H)^h \, (p_L)^{n-h} $$

Note only one binomail coefficient is used. Now imagine $p_H \leftarrow p_H + p_M$.

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okay I see where the coefficient comes from comes from. So now my question is why choose to do n-h choose m? Couldn't we also do n-h choose l? – TheHopefulActuary Jan 17 '13 at 15:18

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