# 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.

• Complement of the event "at least one student AND at least two teachers" is "either there will be no student or there will be at most one teacher. – Dbchatto67 Mar 29 at 5:02

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}

• Is the new equation 1- P(no students) - P(at most 1 teacher) ? Also how would I calculate the at most 1 teacher part? – Manolo Moises Guasco Mar 29 at 5:21
• sorry if this a really dumb question but why is everything added? Shouldn't it be subtraction? – Manolo Moises Guasco Mar 29 at 5:43
• thank you so much! :) – Manolo Moises Guasco Mar 29 at 5:48