tricky probability

pemdas wrote:couldn't get the last one, is it something put in bold?
i'm arranging out of 8, six 6 members except for M A

LalaB wrote:

pemdas wrote: agree about this, we can do so. In new q. we have 8-2=6
we can check M A _ one slot and 6 members, 6C1=6!/5!=6 (fine )

hey pemdas, thnx for ur help. dont u think that there is smth forgotten here. if u change the number of persons to 8, u will have 2 sub-committies with 4(!) ppl, not 3

still, I feel that I was wrong. I have read all solutions above, and I hope I understand them.

I think I misunderstood the q, and wrongly thought, that M cant be without A.

pemdas, in ur new q with 8 ppl, u have 2 sub-committies with 4 ppl, but here "M A _ one slot" u should write "M A _ two slots" ,because now u have 2 sub-c. with 4 (not 3) ppl

anyways, I was wrong, and my method was bad. my fault is that i didnt even get a q.stem.
I thought we need to find out M+A/all, but actually the q stem asked M+A/M+rest (rest=all-M)

yes, you're right, i've edited solution

We need to find the subcommittees of three members that include Michael, and the subcommittee which include Michael and Anthony as well.

Hi pemdas,
Thanks for your reply.

I'm still confused about one thing. Aren't we supposed to find the total no of sub committees that have Michael and Anthony wrt to the total no of all possible 3 member sub committees? ie 4/6C3 ways?
Why are we trying to find out the subcommittees of three members that include Michael at the denominator??

urshohini,
I think You are getting the question wrong. When I first read question, I was thinking the same way you are thinking.

Pay close attention to these terms: Anthony and Michael sit on the six member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

Question is:
Prob = Three-person sub committees formed by Michael that also include Anthony/(Three-member sub committees formed by Michael)

Let 6 members be M, A, B, C, D, E
M-Michael
A- Anthony

Three-person sub group committees formed by Micheal = {MAB, MAC, MAD, MAE, MBC, MBD, MBE, MCD, MCE, MDE}= 10
Three-person sub group committees formed by Micheal that also contain Anthony = {MAB, MAC, MAD, MAE)= 4

Prob = 4/10 = 40%

Oh yes, now I got it... You are right, I misinterpreted the question!!
Thanks again for bringing this to my notice!