The probability of forming a committee under some condition $X$ is given by
$$ \frac{\text{Number of committees that favor } X}{\text{Total number of committees}} $$
Total number of committees that can be formed from 10 students by selecting 3 students $= \binom{10}{3} = \frac{10}{3}\binom{9}{2} = \frac{10\times 9}{3\times 2}\binom{8}{1}$.
1. If $A$ must be in the committee, total number of such committees = $\binom{9}{2}$.
2. Same as the previous case, $\binom{9}{2}$.
3. If both $A$ and $B$ must be in the committee, then we have freedom only in choosing the other member. Hence, number of such committees $= \binom{8}{1}$.
4. Total number of such committees is just the sum of the number of favorable committees in the previous cases.