1. ## binomial distribution(maybe)

A shipment of 20 parts contains 2 defectives. Two parts are chosen at random from the shipment and checked. Let the random variable Y denote the number of defectives found. Find the probability function of this random variable. Find mean and variance of random variable Y.

2. The population is too small to assume independence.
Once again Y can have three values.
$f(y) = \frac{{{2} \choose {y}} {18 \choose {2 - y}}}{{{20} \choose {2}}}$.

3. Originally Posted by Plato
The population is too small to assume independence.
Once again Y can have three values.
$f(y) = \frac{{{2} \choose {y}} {18 \choose {2 - y}}}{{{20} \choose {2}}}$.
I never seen this formula. Where does the denominator come from, Plato? My prof. said in our course up to now we are going to assume that the probability of an event stays the same, in the sense that if we pick uo an item then the probaiities of the others are not affected(which on the contrary is often the case). Does here the probability is changing?

4. Originally Posted by 0123
My prof. said in our course up to now we are going to assume that the probability of an event stays the same, in the sense that if we pick an item then the probabilities of the others are not affected(which on the contrary is often the case). Does here the probability is changing?
I cannot understand an instructor saying that. Clearly with only twenty items, removing one changes the population sufficiently as to change the probabilities. So I don’t know how to help you.

As to the notation ${ {20} \choose {2}}$ is the combination of twenty items taken two at a time. That is the number of ways to pick two representatives from twenty individuals. If Y=1 we select one defective from two and one good from eighteen.