M23-30

Question

A committee of 3 people must be formed randomly from a group of 6 individuals. If Tom and Mary are both part of this group of 6, what is the probability that Tom will be chosen for the committee, while Mary will not?

A.  \frac{1}{10}  
B.  \frac{1}{5}  
C.  \frac{3}{10}  
D.  \frac{2}{5}  
E.  \frac{1}{2}

minhaz3333 · Accepted Answer

In my way,

I found this to be the  easiest solution  for the question.
 ways of  choosing 3 people out of 6 = 6C3  
 ways of  choosing only Tom from Tom and Mary = 1   
and   choosing remaining 2 from remaining 4 (excluding tom and Mary) = 4C2 
 
Probability of choosing 3 member including Tom but not Mary =
 1 x 4C2
--------- = 3/10
 6C3

Bunuel · Answer

Official Solution:

A committee of 3 people must be formed randomly from a group of 6 individuals. If Tom and Mary are both part of this group of 6, what is the probability that Tom will be chosen for the committee, while Mary will not?

A.  \frac{1}{10}  
B.  \frac{1}{5}  
C.  \frac{3}{10}  
D.  \frac{2}{5}  
E.  \frac{1}{2}

We need to find the probability of having Tom (T), not Mary (NM), and another person who is not Mary (NM) on the committee (T, NM, NM).
 
The probability of this combination (T, NM, NM) can be calculated as follows: 
 
 P(T, NM, NM)=3*\frac{1}{6} * \frac{4}{5} * \frac{3}{4}=\frac{3}{10} 
 
We multiply by 3 because the (T, NM, NM) scenario can occur in 3 different ways: (T, NM, NM), (NM, T, NM), or (NM, NM, T).

Answer: C

manish2014 · Answer

I am getting ans as 1/2

we have to find the probability of getting (T,NM,NM)

NM,NM - 5C2=10

and total ways- 6C3=20

therefore - 10/20 =1/2

Bunuel · Answer

It should be 4C2: choosing 2 out out of 4 (6 - Tom - Mary).

WoundedTiger · Answer

Committee of 3 can be formed out of 6 people in 6C3 ways or 20 ways

Now Consider committee in which Tom is already a member so out of remaining 4 members (excluding Mary)..2 can be selected in 4C2 ways = 6 ways..

So probability 6/20 or 3/10

mvictor · Answer

i solved it this way:
6C3 = 20
combinations to choose Tom = 3C1 = 3
combinations not Mary = 3C1 = 3

we add these two 3+3 = 6
6/20 = 3/10

shasadou · Answer

6!/3!3! = 20 total possible combinations. 5!/3!2! = 10 combinations when the 2 are chosen together. 4!/3!1! = 4 comb when the 2 are out.

hence 1 - 14/20 = 3/10 combinations when either is in the team.

happyface101 · Answer

The solution makes sense, but can someone please let me know what I did wrong?

Total number of ways = 6C3 = 6*5*4/(3*2) = 20 ways -- this represents order doesn't matter

Total number of ways to select Tom, Not Marie, Not Marie = 1*4*3/(3*2) = 2 ways -- divide by 3*2 b/c order doesn't matter (to match denominator)

Thus 2/20 = 1/10; why is this wrong??

chetan2u · Answer

Hi,
 you have gone wrong here.. 
in initial 6C3, 3*2 was there because 3 people were selected and these 3 can be arranged in 3! ways..
But in second case, you have already selected 1 out of 3, and you have to select 2 out of remaining 4..
 these two can be arranged in 2! so you have to divide by 2! and not 3!=3*2..

second case is 6 people- 1 already selected and 1 left out .. 
2 to be picked up from 4..
so 4C2=4!/2!2!=6..
ans 6/20=3/10

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

odd_major · Answer

Hi  Bunuel ,

I generally get stuck in such questions. If I take the P&C route of calculating the answer to this question, I can quickly get a 3/10.

However, when I take the probability-based approach (like the one you shared in the OA), I get 1/10.

Since the committee is not an ordered set, there is no difference between one arrangement and another; therefore, there shouldn't be any need to multiply by 3 in the probability. In my opinion, (T, NM, NM), (NM, T, NM), or (NM, NM, T) are all the same thing, per the question. Using this, I get a 1/10.

However, confusingly, if I apply the same logic and solve it via the P&C way: total ways -> 6C3, favourable ways -> 4C2, I quickly get 3/10.

Could you please help me understand where I always go wrong with such questions?

Bunuel · Answer

You can think about this in two ways:

1. When selecting three people one by one, you can get T, NM, NM in three different ways: {T, NM, NM}; {NM, T, NM}; or {NM, NM, T}. So, you need to account for all these possibilities.

2. The denominator, 6 * 5 * 4, counts ordered triplets (all possible sequences). To stay consistent, the numerator must also account for order. If order were not considered, the denominator would be 6C3 = 20, instead of 6 * 5 * 4 = 120, and the numerator would be 4C2 = 6, instead of 3 * 4 * 3 = 36.

Generally, the key is to ensure consistency between the numerator and the denominator. If one uses ordered groups, the other should as well, and vice versa. In the combinations approach (4C2 / 6C3), both the numerator and the denominator represent unordered groups, maintaining consistency. Similarly, in the probability approach, both the numerator and the denominator account for ordered groups, which also keeps the calculation consistent.

Hope it helps.

flowerbead · Answer

I don’t quite agree with the solution. A comittee selection should be a combinatorial, the order of selection should not matters unless stated explicitly - the solution is wrong.

Bunuel · Answer

The solution is absolutely correct. You just don't get it. You can solve this question using a combinatorial approach and get the same result: P=\frac{C^1_1 * C^2_4}{C^3_6}=\frac{1*6}{20}= \frac{3}{10} . If you're still having trouble grasping the concept, I suggest reviewing the discussion above more carefully and checking another discussion of this question here: https://gmatclub.com/forum/a-committee- ... 81051.html Hope it helps.

M23-30

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