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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