A committee consisting of 6 members is randomly selected from 25 students, 5 teachers, and 10 parents. I wish to find the following:

(i) the probability of having no teacher on the committee

(ii) the probability of having neither students nor parents on the committee.

(iii) the probability of having an all parents committee given that the first two selected members are parents.

(iv) the probability of having at least a teacher or at least a student on the committee

For (i), if no teacher is to be on the committee, then there are 35 people to choose from. So the probability is $$\frac{\binom{35}{6}}{\binom{40}{6}}.$$

For (ii) If we are to have neither parents or students, then the committee cannot be formed, since we would only have 5 teachers to choose from. so the probability is $0$.

For (iii) I got the following: $$ \frac{\binom{10}{6}/\binom{40}{6}}{\binom{10}{2}\cdot \binom{30}{4}/\binom{40}{6}}=\frac{\binom{10}{6}}{\binom{10}{2}\cdot \binom{30}{4}}.$$

Is (iv) equivalent to finding 1 minus the probability of having neither teacher nor student on the committee? If so, then I would get $$1-\frac{\binom{10}{6}}{\binom{40}{6}}.$$

Is what I have done okay? Are there better ways of doing them?


You’ve done fine with (i),(ii), and (iv), but (iii) isn’t right. Having picked two parents, you’re now in the position of choosing a $4$-person committee from a group consisting of $25$ students, $5$ teachers, and $8$ parents. The probability that you pick $4$ parents to fill out the $6$-person committee is $$\frac{\binom84}{\binom{38}4}\;.$$


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