Total possibilities: 6C3
Possibilities with Anthony and Michael in the same subcommitee is: 4

4/20=1/5=20%

Hope it's ok

My random guess is 50 %

IMO C..

6 members and two committees with 3 each.

Michael will be on one of the committees out of 6
So, Anthony has 5 positions to be in. He can be in any of the 5 with equal probability and 2 out of 5 will share it with Anthony

So 2/5 = 40%

do let us know the OA..

RC problem in my first post.

There are 5C2 subcommitees with Michael in.
4 can also have Anthony.

So 4/10 or 40% yep

pepeprepa wrote:RC problem in my first post.

There are 5C2 subcommitees with Michael in.
4 can also have Anthony.

So 4/10 or 40% yep

Why 4 can also have Anthony?

Because in my view, I am not sure about the answer I don't want to mislead you, we have:
M A 1 2 3 4 people

MA1
MA2
MA3
MA4

4 groups in which M and A are together.
These 4 groups are among the 5C2 ones.

What's the OA?

Ye, I-ve got it!! ))
pepeprepa, thank you!
4 from 20 is 20%

I think the answer is 20% as well. Here is how I solved it. I'm not sure of the answer, though.

What we need to find is:
(The probability of having Anthony in sub-committee 1 AND the probability of having Michael in sub-committee 1)
OR
(The probability of having Anthony in sub-committee 2 AND the probability of having Michael in sub-committee 2)

Mathematically, this will translate into:
The probability of having Anthony in sub-committee 1 = count of individual to select/count of available individuals to select from = 1/6
The probability of having Michael in sub-committee 1 = count of individual to select/count of available individuals to select from exculding Anthony = 1/5
The probability of having Michael in sub-committee 2 = count of individual to select/count of available individuals to select from exculding the three individuals already assigned to sub-committtee 1 = 1/3
The probability of having Michael in sub-committee 2 = count of individual to select/count of available individuals to select from exculding the three individuals already assigned to sub-committtee 1 and Anthony = 1/2

Now substitute the numbers above:
(1/6)*(1/5)+(1/3)*(1/2) = (1/30) + (1/6) = (1+5)/30 = 6/30 = 1/5 = 20%

Hi jamesk486..i would be nice if u can post the OA. I request Ian/ Stuart to pitch in and clear the confusion. For me its still 40%.

4meonly
What I did is:
- to count the total number of subcommitte groups we can make with Michael in. If you want we can write it like " M _ _ ". You can see that there are two available seats and there are 5 guys who can come there. So that is why I think the total possibilities are 5C2.
5C2= 5!/(2!3!)=(5*4)/2=10
- to count the number of groups of Michael in which there is also Anthony: 4

Finally, 4/10 or 40%
Hope it is ok

pepeprepa wrote:4meonly
What I did is:
- to count the total number of subcommitte groups we can make with Michael in. If you want we can write it like " M _ _ ". You can see that there are two available seats and there are 5 guys who can come there. So that is why I think the total possibilities are 5C2.
5C2= 5!/(2!3!)=(5*4)/2=10
- to count the number of groups of Michael in which there is also Anthony: 4

Finally, 4/10 or 40%
Hope it is ok

5C2? I've made so:
All possibilities to make group of 3 from 6 is 6C3 = 6!/(3!3!)=20
We can have only 4 grous with M and A, so 4/20=20%

I cant find mistake in my reasoning.

For me, with 6C3 you count the total number of possibilities
With 5C2, you count all the commitees in which you have Michael.

IMO A.