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?
(a). 20%
(b). 30%
(c). 40%
(d). 50%
(e). 60%
OA is C
How can solve this question? An Expert view will be appreciated.
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Since Michael's subcommittee must be composed of 3 people, the 2 remaining slots on Michael's subcommittee must be filled from the 5 remaining people.Roland2rule wrote: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?
(a). 20%
(b). 30%
(c). 40%
(d). 50%
(e). 60%
Thus:
From the 5 remaining people, the probability that Anthony is selected for the 2 remaining slots on Michael's subcommittee = 2/5 = 40%.
The correct answer is C.
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Let's assume that we're creating subcommittees.Roland2rule wrote: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?
(a). 20%
(b). 30%
(c). 40%
(d). 50%
(e). 60%
OA is C
How can solve this question? An Expert view will be appreciated.
We want to place 6 people in the following spaces:
_ _ _ | _ _ _
First, we place Michael in one subcommittee; it makes no difference which one:
M _ _ | _ _ _
Now place Anthony.
We can see that there are 5 spaces remaining. 2 spaces are on the same subcommittee as Michael.
So the probability that they are on the same subcommittee is 2/5 = 40%
In other words, 40% of the possible outcomes feature Anthony and Michael on the SAME committee
Answer: C
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Brent