Some P+C questions + probability (1 Viewer)

[S1M0]

Hey guys, haven't got any answers, so i can't check my answers (got sheets from a yr12 textbook in ext. class). These questions i don't neccessarily get though, so if you can - please help.

A committe of 3 women + 5 men is to be selected at random from 7 women and 7 men.
(a) Find the no. of ways that the committee can be chosen
(b) Among the group is a married couple. Find the probability that they will both be slected.

7. There are 6 books about Ancient Rome on a certain shelf in th library. If there are 20 books on the shelf altoghether and I choose 6 at random, find the probability they will all be about ancient rome.

Thanks.

 

[umop 3pisdn]

(a) 7C5*7C3 = 735
It's times and not plus. There's 7C5 ways of choosing the men, and for each of those there's 7C3 ways of choosing the women. So for each of the 21 possible selections of men, there's 35 selections of women.

(b) Just assume the couple are in the committee. Now there's 2 women and 4 men to select from 6 women and 6 men; same as part (a) now, 6C2*6C4 = 225
So of the 735 possible committees, 225 of them have a particular couple in them.
225/735 = 15/49 or about 0.31

Q7. The ways of choosing 6 books is 20C6 = 38760
This is easy because there's only ONE of those selections that contain all 6 books about ancient rome, so the probability is just 1/38760 or about 2.6x10^-5.

The case where you want "at least one" books on ancient rome or something like that is more difficult

 

[jemsta]

i really forgot my ext 1 maths stuff already

 

[S1M0]

(a) 7C5*7C3 = 735
It's times and not plus. There's 7C5 ways of choosing the men, and for each of those there's 7C3 ways of choosing the women. So for each of the 21 possible selections of men, there's 35 selections of women.

(b) Just assume the couple are in the committee. Now there's 2 women and 4 men to select from 6 women and 6 men; same as part (a) now, 6C2*6C4 = 225
So of the 735 possible committees, 225 of them have a particular couple in them.
225/735 = 15/49 or about 0.31

Q7. The ways of choosing 6 books is 20C6 = 38760
This is easy because there's only ONE of those selections that contain all 6 books about ancient rome, so the probability is just 1/38760 or about 2.6x10^-5.

The case where you want "at least one" books on ancient rome or something like that is more difficult

Heh, great. That means i got q.6 right. Thanks!

And for question 7. Arrrghh! I CANNOT believe it was that easy. I sound so stupid for asking how to do that question.

Oh well, thanks mate. Appreciate it. I always do. You know i do.