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.

The given solution is close but indeed incorrect.
Note that you can also look at it as a sample space that has $30000P2=30000\times29999$ outcomes of which $1000\times29000$ are favorable, so that the probability is:$$\frac{1000\times29000}{30000\times29999}$$
Just think of a "bag" filled with $30000$ students of which $1000$ are marked as selected. You pick two out one by one and the outcome is favorable if the first ($A$) appears to be a selected one and the second ($B$) an unselected one.
