i suppose it shud be ~20%

1 X 4C1 / 6C3 = 20%

Number of all possible subcommittees is (6c3*3c3)/2 = 10

Number of subcommittees that include both michael and anthony
= 4c1*3c3 = 4

Therfore the probability is 4/10 = 2/5.

I think the answer is %30

possible subcommutties are cimbination between 6 and 3=20

In each of these subcommuties,they can sit together in 6 different ways

MA_
M_A
AM_
A_M
_MA
_AM

So possibility of sitting together is 6/20=30/100

Is it wrong?

Ans should be 20%

6C3 =20 = nos of 3 poeple committes that can be formed

I dont think here we need to divide the possible committes into two coz even if that is done while taking the percentage we will need to evaluate it like this

(Nos of favorable combinations of committe 1 + Nos of favorable combinations of committe 2) /20

out of these 4 committes i.e M,A+ 1 comibination with each of the 4 directors left

sp reqd probability = 4/20 = 1/5 =20%

Well the answer is 20%.

The no of different ways in which a 6 member team could be divided into two 3 member teams are: 6C3= 20

Now we have to find out the number of possible combination in which a team can have both A and M
say this team is (AM)X:
Now this X could be any person out the remaining 4. so the number of ways in which team can be formed having both A and M is 4

now the percentage would be: 4/20 X 100= 20%

Thanks
Komal Joshi

Think of a simplified problem: 4 persons forming 2 sub committees of 2 each.

Persons: A, B, C and D

4c2 * 2c2 - Number of total combinations
Comm1 comm2
A B C D
C D A B
---
---

First two combinations are really the same. Hence the number of possibilities is reduced by half.

Hence for 6 persons forming 3 sub committees total number of combinations is 6C3*3C3/2 = 10 combinations.

Combinations that include michael and anthony = 4
Therfore the probability is 4/10, which is 2/5 = 40%.