Calculation on a multinomial question I'm working through the following multinomial question: Thirty items are inspected in a graded quality control procedure. The possible grades that can be awarded to an item are fail, major flaw, minor flaw and perfect. Uncertainty about how many items will achieve each grade is to be modelled with a multinomial distribution, and the expected numbers of items achieving each grade are as follows.
Fail = 2; Major flaw = 8; Minor flaw = 15; Perfect = 5.

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*What is the probability that more than one item fails?

My working out:
$P(x > 1) = 1-P(x\le1)$
Because we're looking at those fails greater than one, I thought it would be easier to find those values less than or equal to one.
My problem - how do I make the multinomial distribution from this?
I thought it would be $$1-\frac{29!}{1!8!15!5!}(\frac{1}{4})(\frac{1}{4})^8(\frac{1}{4})^{15}(\frac{1}{4})^5$$
I chose $n=29$ in this case given that there are only 2 fails and we're looking for 1 or less to find them. In doing so I'd have to minus this from the total number.
However this gives me a very large probability like 0.999 which I do not think is right.
 A: You're told that $\ 2,8,15\ $ and $\ 5\ $ are the expected numbers  rather than the actual numbers of $\ 30\ $ items that will fall into each of the four grades.  This means that each item must have a probability of $\ p_1=\frac{1}{15}\ $ of failing, $\ p_2=\frac{4}{15}\ $  of having a major flaw, $\ p_3=\frac{1}{2}\ $ of having a minor flaw, and $\ p_4=\frac{1}{6}\ $ of being perfect. The actual numbers $\ n_1, n_2, n_3, n_4\ $ of items out of $\ n\ \ \big($ i.e. with $\ n=n_1+n_2+n_3+n_4\ \big)$ that will fall into each of the four grades are random variables with a multinomial distribution.  That is, the probability of these numbers actually occurring is
$$
\frac{n!}{n_1!n_2!n_3!n_4!}p_1^{n_1}p_2^{n_2}p_3^{n_3}p_4^{n_4}=\frac{n!}{n_1!n_2!n_3!n_4!}\frac{1}{2^{n_3+n_4-2n_2}3^{n_1+n_2+n_4}5^{n_1+n_2}}\ .
$$
The number of fails, by itself, however will be binomially distributed:
\begin{align}
P\big(\text{number of fails }=n_1\big)&=\frac{n!}{n_ 1!(n-n_1)!}p_1^{n_1}(1-p_1)^{n-n_1}\\
&=\frac{n!}{n_ 1!(n-n_1)!}\frac{14^{n-n_1}}{15^n}\ .
\end{align}
For $\ n=30\ $ this becomes
$$
P\big(\text{number of fails }=n_1\big)=\frac{30!}{n_ 1!(30-n_1)!}\frac{14^{30-n_1}}{15^{30}}\ .
$$
The probability of zero fails is therefore $\ \Big(\frac{14}{15}\Big)^{30}\ $ and the probability that exactly one item fails is $\ 30\Big(\frac{14^{29}}{15^{30}}\Big)\ $, so the probability that more than one item fails is
$$
1-\Big(\frac{14}{15}\Big)^{30}-30\Big(\frac{14^{29}}{15^{30}}\Big)=1-\frac{44\cdot14^{29}}{15^{30}}\approx0.6\ .
$$
