1. Good post? |

## Probability

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%

2. Good post? |
IMO C

8/6c3=40&#37;

3. Good post? |
This is my logic please let me know if I am doing anything wrong.

number of ways we can have '2' subgroup is 6C3/2 = 10.

let michael belong in any of this subgroup.

now we try to find for the subgroup where michael belong, in how many ways can we select rest of the two persons so that none of them are Anthony.

this can be done in 4c2 way or 6 way , so total percentage of all the possible sugroups where Michael belongs but anthony does not = 6/10 = 60&#37;

hence michael and anthony both belong in 100-60 = 40% way.

thanks.

4. Good post? |
Thanks cubicle, simply faaaaabulous explanation

5. Good post? |
let the members be 1,2,3,4,M(ichael),A(ntony)..
total no. of ways to select 3 memb. com. 6c3 X 1 (the other 3 will get selected by default once we have the first 3)

let the com. be X & Y.

X can have M & A together as...
M A 1
M A 2
M A 3
M A 4 = 4 ways, similarly com. Y can be selected as well, thus total ways 8

prob. = 8/6c3

6. Good post? |
yes imo c. 4c1/(6c3/2!)

7. Good post? |
Let the groups be XYM & WZA.

Now keeping M fixed in group 1, there are 5C3=10 ways to select member of group 2.

Now lets find the number of ways the members in group 2 can be arranged with "A" always present
4C2 = 6
So the probability that A & B will belong to different groups is 6/10=3/5
So the ans is 1-3/5=2/5= 40&#37; (C)

8. Good post? |
The probability of Of antony in subgroup one is 1/2... then the probability of micheal with him is 2/5 (favorable places / total places)... when we multiply we obtain 1/5.. this is for one sub group.. hence multiplying by 2 we get 40&#37;... por favor, el seņor let me know if I am wrong...

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