<s>[INST] Hypergeometric distribution exercise! A store has $20$ guitars in stock but 3 are defective. Claire buys $5$ guitars from this lot.
(a) Find the probability that Claire bought $2$ defective guitars.
I use $N=20,n=5, k = 3,x=2$ where $N$ is the total sample space, $n$ is the number of trials, $k$ is the number of defectives and $x$ is the number of guitars bought that are defective
Hence I got $h(2;20,5,3) = \frac{{3 \choose 2}{17 \choose 3}}{20 \choose 5} = 0.1316$
Is my answer correct? I think I may be missing something.
 [/INST] We assume that Claire's selection method was such that all collections of $5$ guitars were equally likely to be bought.
There are $\binom{20}{5}$ ways to choose $5$ guitars from $20$.
There are $\binom{3}{2}$ ways to choose $2$ defectives from $3$. For each of these ways, there are $\binom{17}{3}$ ways to choose $3$ non-defectives to accompany them, for a total of $\binom{3}{2}\binom{17}{3}$.
So your expression for the probability is right. 
Now let us compute. The  numerator is 
$3\cdot \frac{(17)(16)(15)}{6}$. This is $2040$. The denominator is 
$\frac{(20)(19)(18)(17)(16)}{120}$, which is $15504$. Divide. Your answer is correct to $4$ significant figures.
A simplified fractional form is $\frac{5}{38}$. 
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