A shipment of 20 similar laptops to a retail outlet contains 3 that are defective. If a school makes a random purchase of 2 of these computers, find the probability distribution for the no. of defectives.
Let $X$ be a random variable whose values $x$ are the possible numbers of defective computers purchased by the shool. Then $x$ can only take the numbers $0$, $1$, and $2$.
$$f(0)= P(X=0)= \frac{\dbinom 3 0 \dbinom{17} 2 }{\dbinom{20} 2}= \frac{68}{95}$$
$$f(1)= P(X=1)= \frac{\dbinom 3 1 \dbinom{17} 1}{\dbinom{20} 2}= \frac{51}{190}$$
$$f(2)= P(X=2)= \frac{\dbinom 3 2 \dbinom{17} 0}{\dbinom{20} 2} = \frac{3}{190}$$
This is an example on one of the slides from the probability course that I'm taking. But it just does not make sense to me. Can anyone please explain it to mt? Thank you.
Best Answer
$X$ follows a hypergeometric distribution. A classic phrasing is that there are $N$ total items, with $G$ good ones, $B = N-G$ bad ones, and $n$ is the number of draws from the items without replacement. If we let $X$ follow this distribution, then the probability that $X = k$ is $$P(X = k) = \frac{\binom{N}{k}\binom{N-G}{n-k}}{\binom{N}{n}},$$
where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, which is the binomial coefficient. Notice that this is not $(n/k)$, which is just regular division. Basically, we are counting all the ways we can get $k$ goods, $n-k$ bads, and dividing by the number of ways we can get $n$.
Notice that in your problem, we are interested in the number of defective items that we draw from the shipment. It is implied that the draws are without replacement. Since we are interested in the defective ones, we will call these the "good" ones. So, $N = 20$, $G = 3$, $B = 17$, $n = 2$. For example, if $k = 0$, then we want to calculate the probability that we draw no defective items: $$f(0) = P(X = 0) = \frac{\binom{3}{0}\binom{17}{2}}{\binom{20}{2}} = \frac{68}{95}.$$
In words, we are counting the number of ways to choose 0 of 3 defective ones, multiplying by the number of ways to choose 2 of 17 working items, and then dividing by the number of ways to draw 2 items from the entire shipment.
You can try the formula for the other values.