A 5 member committee is to be chosen from 15 students and 10 teachers 
A $5$ member committee is to be chosen from $15$ students and $10$ teachers.
  a) Determine the probability the committee will have at least one student AND at least two teachers.

So what I know for sure is that it will be easier to use the indirect method. Therefore,$$1-P(\text{No students})-P(\text{No teachers})=1-\frac{\binom{10}{5}}{\binom{25}{5}}-\frac{\binom{15}{5}}{\binom{25}{5}} \dots$$I know that the no teachers covers for the probability of at least one teacher but I'm not sure where to go from there.
 A: Let $A$ denote the event that there will be no student in the committee and $B$ denote the event that there will be at most $1$ teacher in the committee. Then the required event is $(A \cup B)^c.$
Let $C$ denote the event there will be no teacher in the committee and $D$ denote the event that there will be exactly one teacher in the committee. Then $B = C \cup D.$ Observe that $C$ and $D$ are mutually exclusive events. Hence $\Bbb P(B) = \Bbb P(C) + \Bbb P(D).$ So what is $\Bbb P(A \cap B)$?


$$\begin{align} \Bbb P(A \cup B) & = \Bbb P(A) + \Bbb P(B) - \Bbb P(A \cap B). \\ & = \Bbb P(A) + \Bbb P(C) + \Bbb P(D) - \Bbb P ((A \cap C) \cup (A \cap D)).\\ & = \Bbb P(A)+ \Bbb P(C) + \Bbb P(D) - \Bbb P(A \cap C) - \Bbb P(A \cap D). \end{align}$$


But $A \cap C$ and $A \cap D$ are impossible events. Can you see why? So $\Bbb P(A \cap C) = \Bbb P(A \cap D) = 0.$
Therefore


$$\begin{align} \Bbb P(A \cup B) & = \Bbb P(A) + \Bbb P(C) + \Bbb P (D). \\ & = \frac {\binom {10} {5}} {\binom {25} {5}} + \frac {\binom {15} {5}} {\binom {25} {5}} + \frac {\binom {15} {4} \cdot \binom {10} {1}} {\binom {25} {5}}. \\ & = \frac {7} {22}.\end{align}$$


So the probability of the required event $(A \cup B)^c$ is 


$$\begin{align} \Bbb P((A \cup B)^c) & = 1 - \Bbb P(A \cup B).\\ & = 1 - \frac {7} {22}. \\ & = \frac {15} {22}. \end{align}$$


