probability

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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%


Ans 40%

[logitech]

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%


Ans 40%

For the first group

6c3=20 and for the second group 3c3=1

20x1 = 20 different ways of splitting into two.

if M and A are in the first group then we can have 4 ways of finding the 3rd person

if they are in the second group, 4 ways of finding the 3rd group

(4+4)/20 = 40 %

Any other approaches ?

[Mani_mba]

Approached in the same way as logitech did..

Ans - 40%

[cramya]

Hi Sophia,
The approach suggested by Logitech is good. If u ever get stuck figuring out the formulas list out the possibilities(cant do this for all problems especially with large numbers but it works here).

It's a combination problem so order does not matter(i.e michael anthony bryan is the same as bryan michael anthony etc..).

6 member - > M A B C D E

No of commitees that can be formed with M(Michael)

MAB,MAC,MAD,MAE,MBC,MBD,MBE,MCD,MCE,MDE

U will see that Anthony(A) will be in 4/10 i.e 40% of those subcommitees Michael(M) will be in

[logitech]

Hi Sophia,
The approach suggested by Logitech is good. If u ever get stuck figuring out the formulas list out the possibilities(cant do this for all problems especially with large numbers but it works here).

It's a combination problem so order does not matter(i.e michael anthony bryan is the same as bryan michael anthony etc..).

6 member - > M A B C D E

No of commitees that can be formed with M(Michael)

MAB,MAC,MAD,MAE,MBC,MBD,MBE,MCD,MCE,MDE

U will see that Anthony(A) will be in 4/10 i.e 40% of those subcommitees Michael(M) will be in

I like this discussion!

M X X , so there are two openings and 5 members 5c2=10 ways - total

M A X , 4C1 = 4 ways

4/10 = 40 %

[cramya]

U got it!

[schakiiieee]

Could we also interpret the question like the following?

2 positions remaining in team A, 3 positions in team B. It is like having 2 apples and 3 oranges in a bag - what is the probability of getting an apple by pointing randomly at one of the fruits? 2 / (2+3) = 40%.

Anything wrong with that reasoning?