A committee of 3 has to be formed randomly from a group of 6 people.

Sample space for selecing committe of 3 people out of 6 6C3
Number of way, both TOM AND mary are excluded and committe can be formed 4C3
Number of way, when both are there in teh committe 4C1,
So probability when either of them are there are not= (6C4-(4C3+4C1))/6C3 =60%
In case, only one has to be there, 60%/2 = 30%

IMO 3 / 10

4C2 / 6C3

Another way to look at this is:

We have three units - Tom (T), Mary (M) & the Rest (R).

Total ways of making a community of 3 from 6 people = 6 x 5 x 4 = 120

Now ways the community can be formed with Tom in it and Mary NOT in it:
TRR
RTR
RRT

TRR posb = 1 (way of picking Tom) x 4 (4 people left excluding Tom and Mary) x 3 (3 people left after excluding Tom, Mary and the person selected before) = 12
RTR and RRT have the possibility as TRR. Therefore possibilities the community can be formed with Tom in it and Mary NOT in it = 12 + 12 + 12 = 36

Hence probability \(= \frac{36}{120} = \frac{3}{10}\)

Cheers!
JT

A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?

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


Look for any of the three variants: Tom-notMary-notMary, notMary-Tom-notMary, notMary-notMary-Tom. The probability of one such variant is \(\frac{1}{6} * \frac{4}{5} * \frac{3}{4} = \frac{1}{10}\) . The answer to the question is \(3*\frac{1}{10} = \frac{3}{10}\) .


Why can't this be 5C3/ 6C3 = 1/2? Where did I make mistake when I applied combination formula approach? I know that this leads to wrong answer.

Here is why 5C3 is wrong:
5C3 is the total number of ways to select 3 people from 5. If Tom is one of the 5 people, 5C3 does not count ONLY those ways in which Tom will be included. It counts the number of ways in which 3 people can be selected from 5 people (including Tom). This will include selections which do not include Tom, and so overstate the number of ways of always including Tom in the selection.


The correct solution:
Tom will always be selected, so select the other 2 people from the remaining 4 people. Number of ways = 1 * 4C2 = 6
Total number of ways to select 3 people from 6 = 6C3 = 20
Therefore probability = 6/20 = 3/10

You can solve this problem with direct probability approach as well. We need the probability of T, NM, NM (Tom, not Mary, not Mary). \(P(T, NM, NM)=3*\frac{1}{6} * \frac{4}{5} * \frac{3}{4}=\frac{3}{10}\), we are multiplying by 3 since {T, NM, NM} scenario can occur in 3 different ways: {T, NM, NM}, {NM, T, NM}, {NM, NM, T}.

Answer: C.

A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but 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 T, NM, NM (Tom, not Mary, not Mary).

\(P(T, NM, NM)=3*\frac{1}{6} * \frac{4}{5} * \frac{3}{4}=\frac{3}{10}\), we are multiplying by 3 since {T, NM, NM} scenario can occur in 3 different ways: {T, NM, NM}, {NM, T, NM}, or {NM, NM, T}.

Answer: C.

For anyone who's still confused, here's some more clarification:

For the subcommittee, you basically want to choose Michael, then choose 2 people who do not include Mary. Order matters - meaning Michael in the 1st slot should be counted separately from Michael in the 2nd and 3rd slots.


(Out of 1 available Michael, pick that 1 Michael) * (Out of remaining 4 people excluding Mary, pick 2 people for the 2nd and 3rd slots)
= -----------------------------------------------------------------------------------------------------------------------------------------------------------------
(Out of 6 available people pick 3 for the committee)

= (1C1) * (4C2) / (6C3)

= (1) * (4*3 / 2) / (6*5*4 / 3!)
= 6 / 20

= 3 / 10 = 30%

This is how I solved it:
There are 3 positions to be filled with 6 people :
Pos 1: Probability that Tom has to be INCLUDED - 1/6
Pos 2: Probability that Mary has to be EXCLUDED - 4/5 (out of 5 remaining people 4 people can be taken in 4 ways )
Pos 3 : Probability that one out of the remaining fill the 3rd position : 3/4
Now the 3 positions are interchangeable : Therefore X3
Solution: (1/6*4/5*3/4)*3 = 3/10

The total number of committees of 3 people from a group of 6 people that can be formed is 6C3 = (6 x 5 x 4)/(3 x 2) = 20. The number of committees can be formed in which Tom will be selected and Mary will not consists of Tom’s being chosen for a spot, and 2 of 4 individuals (but not Mary) chosen for the remaining 2 spots:

1C1 x 1C0 x 4C2 = 1 x 1 x (4 x 3)/2 = 6

Therefore, the probability that Tom will be selected for the committee but Mary will not is 6/20 = 3/10.

Answer: C

Hi experts,

Following is my solution. Please help me to correct me because I couldn't find what is wrong with my way. Many thanks

+ Total number of ways to form a group of 3: 6C3 = 20
+ Number of ways to form a group of 3 which have Tom, but not Mary: 1*4*3 = 12
=> Probability: 12/20 = 3/5

Brunel, when we use the combinatory approach, why do not we count all the arrangements here?

I expected to calculate the total cases as C(6,3) 3! not as C(6,3).

Here's how I thoguht about it.

A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?


A. 1/10

B. 1/5

C. 3/10

D. 2/5

E. 1/5

1/6 x 4/5 x 3/4 = 3/30 <--- P(Tom) x P(anybody but Mary) x P(anybody but Mary less 1 person)

3/30 x 3 = 9 /30 = 3/10 <-- Tom can come first, second, or third.

C

6 people and 3 spots (= x)

x x x y y y

Probability Tom will be one of the x: 3/6.

Two spots left:

x x y y y

Probability Mary will be one of the y: 3/5.

1/2 * 3/5 = 3/10

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