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

there are many possiblities events to get exactly one student on the committee, are (1S, 5T),(1S, 5B),(1S, 1T,4B),(1S,2T,3B),(1S,3T,2B),(1S,4T,1B)
so, the probability =
(8C1*10C5+8C1*6C5+8C1*10C1*6C4+8C1*10C2*6C3+8C1*10C3*6C2+8C1*10C4*6C1) divided by 24C6

Answer 2

For a), it is simply:

8(16C5)/(24C6)

Because you can look at a person as a student(8) or a non-student(16)

Answer 3

Similarly, for b) you can look at a person as a businessman(6) or a non-businessman(18), so:

(18C6)/(24C6)

is the answer.

Answer 4

For c), you can look at a person as a teacher(10) or a non-teacher(14), so you have:

1 - (14C6)/(24C6)

Answer 5

hmmmmm... i dont think no1) similar by my job, but is it different with my job @tcarroll010

Answer 6

I didn't take time to look if our different methods gave the same answer, but I know that if you look at my method of "category" and "non-category", it will give the right answer and it does so with very little computation.

Answer 7

For a), you are looking at the number of ways to get a sub-committee of 5 from all available non-students. You then multiply that by a pick of one of any of 8 students. That is all divided by the number of ways to pick any committee of 6 from 24.

Answer 8

For b), 18C6 is the way to get a committee of 6 from all available non-businessmen. Again divided by the number of ways to pick any committee of 6 from 24.

Answer 9

c) is very similar to b) except that firstly, we are dealing with teachers and non-teachers, and secondly, we have "1 minus a probability" because we are subtracting "no teachers at all". So, 1 minus a probability gives "at least one teacher"

Answer 10

after I think about it again, u are right @tcarroll010 ... exactly, it is a simple problem. but it will same with my job, but too long :)

Answer 11

It is good you see that. To do a) or any of the problems for that matter, we don't have to bog ourselves down with specifically what type of person every committee member is, just whether "student or non-student" or "businessman or non-businessman" etc.

Answer 12

You would have to consider all categories only if you are asked to find probabilities where the committee consists of all 3 types.

Answer 13

thnx so much guys for explain to me that's so helpful

Answer 14

Probability questions are always very interesting. It was great working with both of you folks. You're welcome!

Answer 15

yes, i gree that... thanks, @tcarroll010 for ur opinions :)

Answer 16

*agree

Answer 17

And thanks for a good, solid, alternate solution. @RadEn