Probability that one person is selected and other is not selected in a group There are $30000$ students, out of which only $1000$ students are selected. There are two students $A$ and $B$, what is the probability that $A$ is selected and $B$ is not selected?
The solution is given as $\frac{1000}{30000}* \frac{29000}{30000}$, but this does not seems correct to me.
The way I look at this problem, the sample space has $_{30000}C_{1000}$ elements.
The event, since we need all the sets from $SS$ that has $A$ but not $B$, we have $_{29998}C_{999}$
Since we fix $A$ in the set and exclude $B$
Both give very different results, request your help in understanding which one is correct and why, Thanks.

Your answer is correct, and theirs is very slightly incorrect. Letting $A$ represent the event where student A is selected and $B$ represent the event where student B is not selected, their solution is based on the logic that
$$P(A \cap B) = P(A)P(B)$$
which is the multiplication rule for independent events. However, whenever you sample a population without replacement, the selections are dependent events, so the proper logic would actually be to use the general
$$P(A \cap B) = P(A)P(B \ \vert\  A)$$
which follows from the definition of conditional probability. In our case, $P(B \ \vert \ A)$ is the probability of $B$ not being selected given that $A$ was already selected, which ends up being $\dfrac{29000}{29999},$ essentially because we need to subtract $A$ from the $30000$ students originally under consideration, and the numerator stays the same because exactly the same number still need to be passed over.
So the correct answer would be $\dfrac{1000 \cdot 29000}{30000 \cdot 29999},$ which ends up being the same as your answer:
$$\frac{29998\choose999}{30000\choose1000} = \frac{\frac{29998!}{999!28999!}}{\frac{30000!}{1000!29000!}} = \frac{1000!}{999!} \cdot \frac{29000!}{28999!} \cdot\frac{29998!}{30000!} = \frac{1000 \cdot 29000}{30000 \cdot 29999}$$
