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Probability - MGAMT

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Probability - MGAMT

by singh_amit19 » Sat Oct 06, 2007 12:25 am
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?


20%
30%
40%
50%
60%
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by singh_amit19 » Sat Oct 06, 2007 12:30 am
i suppose it shud be ~20%

1 X 4C1 / 6C3 = 20%
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by jay2007 » Mon Oct 08, 2007 12:27 pm
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.
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by emrahinci » Tue Oct 09, 2007 12:05 am
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?
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by samirpandeyit62 » Wed Oct 10, 2007 4:18 am
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%
Last edited by samirpandeyit62 on Thu Oct 11, 2007 2:46 am, edited 1 time in total.
Regards
Samir
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by joshi.komal » Wed Oct 10, 2007 1:24 pm
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
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by jay2007 » Wed Oct 10, 2007 7:54 pm
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%.
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