Probability Calculation using combinations In a population of $250$ items, $20$ are defective. Suppose $4$ items are sampled at random, without replacement. 
a. What is the probability that the sample will consist of $4$ defective items?
Solution (don't know if it is correct or not): $P$($4$ defective) = $\dfrac{1}{\binom{20}{4}} = \dfrac{1}{4845}$
b. What is the probability that the sample will consist of $3$ or fewer defective items?
Solution (again, don't know if correct or not): $\dfrac{1}{\binom{20}{3}} + \dfrac{1}{\binom{20}{2}} + \dfrac{1}{\binom{20}{1}}$
c. What is the probability that the sample will consist of neither zero nor four defectives? (NO clue how to do this one)
Help would be greatly appreciated. Thank you. 
 A: First Question: There are $\binom{250}{4}$ ways to choose $4$ items from $250$. All these choices  are equally likely. 
There are $\binom{20}{4}$ ways to choose $4$ defectives. So the required probability is 
$$\frac{\binom{20}{4}}{\binom{250}{4}}.$$
Second Question: The answer is most easily found as $1$ minus the answer to the first. 
Third Question: The simplest way is to first find the probability of $0$ or $4$ bad. (This is the complement of the event we are interested in.) The probability of $4$ bad has already been found. The probability of $4$ good is found in the same way, except that in the numerator we have $\binom{230}{4}$. So the probability of neither $0$ nor $4$ is 
$$1-\frac{\binom{20}{4}}{\binom{250}{4}}-\frac{\binom{230}{4}}{\binom{250}{4}}.$$
Remark: There are easier ways to solve these particular problems. We chose the approach ABOVE because it generalizes nicely to more complicated problems. But let's solve the first problem a simple way. 
Imagine choosing one at a time. The probability the first is bad is $\frac{20}{250}$. Given the first was bad, the probability the second is bad is $\frac{19}{249}$. And given the first two are bad, the probability the third is bad is $\frac{18}{248}$. Continue. But there is only one more step. So the probability they are all bad is
$$\frac{20}{250}\cdot\frac{19}{249}\cdot\frac{18}{248}\cdot\frac{17}{247}.$$
A: You employ Hypergeometric Distribution for such problems:
a) $$\dfrac{\binom{230}{0} \binom{20}{4}}{\binom{250}{4}}$$
b) $$1-\dfrac{\binom{230}{0} \binom{20}{4}}{\binom{250}{4}}$$
c) $$1-\dfrac{\binom{230}{0} \binom{20}{4}}{\binom{250}{4}}-\dfrac{\binom{230}{4} \binom{20}{0}}{\binom{250}{4}}$$
