Question

A committee of 6 people is chosen from 8 students, 10 teachers, and 6 business
executives. What is the probability that:
a) there is exactly one student on the committee
b) there are no business executives on the committee
c) there is at least one teacher on the committee

Answer 1

i'll you try

Answer 2

one out of 8, the other 5 can be any of the 16 remaining
numerator is \(\binom{8}{1}\times \binom{1}{5}\) denominator is \(\binom{24}{6}\)

Answer 3

*let*

Answer 4

oops big typo there

Answer 5

\[\binom{8}{1}\times \binom{16}{5}\]

Answer 6

OMG THE GREAT SATELLITE. I've seen you so much when I was looking for help, thanks a lot. That's exactly what I had for the answer to a) too. And sourwing, I'm done, this is just to see if I'm correct or not ;)

Answer 7

Ok so I got 34944 for the answer, the only thing that was messing me up here is the denominator, how do I find it?

Answer 8

there are 24 people and you're choosing 6

Answer 9

OH YES \[\left(\begin{matrix}24 \\ 6\end{matrix}\right)\] THANKS AGAIN SOUR

Answer 10

yes, you can add up the probabilities that have at least 1 teacher. Or you can find the probability that contain NO teacher and subtract from the universal set

Answer 11

No businessmen would require missing the businessmen in all six picks, which is (18/24)(17/23)(16/22)(15/21)(14/20)(13/19).

The problem of at least one teacher can be handled as @sourwing just noted after doing this kind of calculation for the no teacher situation.

I am not familiar with the notations used above to explain the probabilities, so my comment may be equivalent to theirs...or even wrong.

Answer 12

That is also correct douglas, but just so you know, the above notation is for Combinations, where order of the people is not important. \[\left(\begin{matrix}24 \\ 6\end{matrix}\right)\] is 24C6 which means 6 people are being chosen from a total of 24.

Answer 13

Thank you for explaining it. Regards.