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# Anthony and Michael sit on the six-member board of directors

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Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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15 Jul 2013, 13:07
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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?

Probability approach:

(AMX XXX) or (XXX AMX)

AM could be in the first committee or in the second committee.
When AM is in the first committee, we select A first and then the probability of selecting M between the remaining 5 persons is 1/5.

The same for the second committee.
Then:
$$P = \frac{6}{6} * \frac{1}{5} + \frac{6}{6}*\frac{1}{5} = \frac{12}{30} = 40%$$

Reversal probability approach:
AMX XXX
q= probability of A and M to be in different committees.
P = 1 - q = probability A and M together in the same committee.

$$q = \frac{2}{2}*\frac{4}{5}*\frac{3}{4} = \frac{12}{20} = 60%$$

For any of the committes:
$$\frac{2}{2}$$ we select A or M from A or M.
$$\frac{4}{5}$$ probability of picking any of the Xs from the remaining XXXXM
$$\frac{3}{4}$$ probability of picking another X from the remaining XXXM

then P = 1 - q = 1 - 60% = 40%

Combinatorial approach:
AMX XXX

We select the group (AM) from the 2 groups (AM) (XXXX). Then we combine the group AM with each X of the group XXXX:

$$P = \frac{{C^2_1 * C^4_1}}{{C^6_3}} = \frac{{2*4}}{20} = 40%$$

Reversal combinatorial approach:

q= probability of A and M to be in different committees.
P = 1 - q = probability that A and M being together in the same committee.

AM XXXX

We choose A from (AM) $$(C^2_1)$$ and then 2 more X (XX) from the remaining (XXXX) $$(C^4_2)$$:

$$q = \frac{{C^2_1 * C^4_2}}{{C^6_3}}= \frac{{12}}{20} = \frac{12}{20} = 60%$$

P = 1 - q = 1 - 60% = 40%
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Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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15 Jul 2013, 13:14
Maxirosario2012 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?

Probability approach:

(AMX XXX) or (XXX AMX)

AM could be in the first committee or in the second committee.
When AM is in the first committee, we select A first and then the probability of selecting M between the remaining 5 persons is 1/5.

The same for the second committee.
Then:
$$P = \frac{6}{6} * \frac{1}{5} + \frac{6}{6}*\frac{1}{5} = \frac{12}{30} = 40%$$

Reversal probability approach:
AMX XXX
q= probability of A and M to be in different committees.
P = 1 - q = probability A and M together in the same committee.

$$q = \frac{2}{2}*\frac{4}{5}*\frac{3}{4} = \frac{12}{20} = 60%$$

For any of the committes:
$$\frac{2}{2}$$ we select A or M from A or M.
$$\frac{4}{5}$$ probability of picking any of the Xs from the remaining XXXXM
$$\frac{3}{4}$$ probability of picking another X from the remaining XXXM

then P = 1 - q = 1 - 60% = 40%

Combinatorial approach:
AMX XXX

We select the group (AM) from the 2 groups (AM) (XXXX). Then we combine the group AM with each X of the group XXXX:

$$P = \frac{{C^2_1 * C^4_1}}{{C^6_3}} = \frac{{2*4}}{20} = 40%$$

Reversal combinatorial approach:

q= probability of A and M to be in different committees.
P = 1 - q = probability that A and M being together in the same committee.

AM XXXX

We choose A from (AM) $$(C^2_1)$$ and then 2 more X (XX) from the remaining (XXXX) $$(C^4_2)$$:

$$q = \frac{{C^2_1 * C^4_2}}{{C^6_3}}= \frac{{12}}{20} = \frac{12}{20} = 60%$$

P = 1 - q = 1 - 60% = 40%

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%

First approach:
Let's take the group with Michael: there is a place for two other members and one of them should be taken by Anthony, as there are total of 5 people left, hence there is probability of 2/5=40%.

Second approach:
Again in Michael's group 2 places are left, # of selections of 2 out of 5 5C2=10 - total # of outcomes.
Select Anthony - 1C1=1, select any third member out of 4 - 4C1=4, total # =1C1*4C1=4 - total # of winning outcomes.
P=# of winning outcomes/# of outcomes=4/10=40%

Third approach:
Michael's group:
Select Anthony as a second member out of 5 - 1/5 and any other as a third one out of 4 left 4/4, total=1/5*4/4=1/5;
Select any member but Anthony as second member out of 5 - 4/5 and Anthony as a third out of 4 left 1/4, total=4/5*1/4=1/5;
Sum=1/5+1/5=2/5=40%

Fourth approach:
Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10
# of groups with Michael and Anthony: 1C1*1C1*4C1=4
P=4/10=40%

OPEN DISCUSSION OF THIS QUESTION IS HERE: anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html
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Re: Anthony and Michael sit on the six-member board of directors   [#permalink] 15 Jul 2013, 13:14

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# Anthony and Michael sit on the six-member board of directors

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