A shipment of 8 similar microcomputers 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 number of defectives.
Q: Is there a way to solve the above with Poisson distribution?
The Poisson distribution is neither called for nor effective here. The cases can be counted directly: 0, 1 or 2 computers will be defective. Labelling all the computers distinctly we have $\binom82=28$ choices.
For 0 defective computers we select any two of the five working computers: $\binom52=10$. For 2 we have three to select from: $\binom32=3$. By exclusion, the remaining $28-10-3=15$ choices have exactly 1 defective.
Thus there is a $\{3/28,15/28,5/14\}$ chance that $\{0,1,2\}$ computers will be defective – the desired probability distribution.