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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.

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    $\begingroup$ 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. $\endgroup$ – Dbchatto67 Mar 29 at 5:02
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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

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

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

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