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.
Let $x$ be a random variable whose values $x$ are the possible numbers of
defective computers purchased by the school. Then $x$ can be any of the
numbers $0,1$ and $2$.
Can I solve this question using binomial distribution?
No, the binomial distribution doesn't apply in this case.
While it's true the values of $X$ are $0,1$ and $2$ the probability of selecting a defective computer depends on what's already been selected. This breaks the assumption of constant probability for the selection event required for the binomial distribution. The selections are not $i.i.d.$ i.e they are not identical and independently distributed.
Before any selection, the probability of selecting a defective computer is $3/8$. Say you then select a defective computer, then there are 2 defectives left and 7 computers. The new probability of a defective is $2/7$.
Of course, you may get a non-defective first time. Then there are still 7 computers left, but now 3 defectives remaining. The probability of a defective on the next selection is then $3/7$.
So the probabilities change, depending on what's already happened, i.e. the events are not independent.