vishalrastogi wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.
The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.
Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.
(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.
Answer: E.
Hope it's clear.
Hello Bunuel
Question Asks us to find the Number of Committees, You have given you solutn by finding the probability.
Although doest makes much difference !
I tried a slight different method. Plz tell me where i am going wrong ?
Sorry for wasting your time
The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.
Let t= x+y.
t shows the total number of members other than the one who are both Managers & Board Members.
We have to find out: How many committees can be formed such that at least one committee member is both a manager and a board member.
So it can be := \(C^1_z . C^1_t + C^2_z\)
where \(C^1_z . C^1_t\) = one member from the members who are both Director & board member, Other from the remaining i.e. t
\(C^2_z\) = both memebers from the members who are both Director & board member.
(1)
The number of committees that can be formed such that neither committee member is both a manager and a board member is 30
t represents the total of neither
so we get : \(C^2_t = 30\) --> \(\frac{(t-1)t}{2}=30\)
Not sufficient.
(2)
The number of committees that can be formed such that each committee member is both a manager and a board member is 10
--> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.
(Same as yours)
1+2)
We had to find : \(C^1_z . C^1_t + C^2_z\)
From eq.2 We know Z
from eq.1 It seems we can find out t - although its not giving integer values. I dont knw why.
This way i was getting C as the answer
Please help
Thanks a lot Again !!