The six male board members of a certain company are outnumbe

Question

The six male board members of a certain company are outnumbered by the female board members. If three board members are chosen at random, the probability that the first two chosen are of the same sex is 1/2. What is the probability that all 3 chosen members are men?

A. 1/77
B. 1/28
C. 1/24
D. 1/21
E. 5/21

Show Answer

Official Answer

B

A. 1/77
B. 1/28
C. 1/24
D. 1/21
E. 5/21

Xen0n · Answer

Step-by-Step Solution:
Define the variables:
Number of men (M) = 6

Let the number of women be F. According to the problem, F > 6.

Total number of members (N) = 6 + F.

Find the number of women (F):The probability that the first two members are of the same sex (either both men or both women) is 1/2.
P(Same Sex) = P(MM) + P(FF) = 1/2

Using conditional probability without replacement:  +   = 1/2  /   = 1/2

Since N = F + 6,

the equation becomes:  /   = 1/2

Cross-multiplying to solve:2 * (F^2 - F + 30) = (F+6)(F+5)2F^2 - 2F + 60 = F^2 + 11F + 30
Rearranging into a standard quadratic equation:F^2 - 13F + 30 = 0(F-10)(F-3) = 0
This gives two possible values for F: 3 or 10. Since women outnumber men (F > 6), we choose F = 10.

Therefore, total members (N) = 6 + 10 = 16.Calculate the probability that all 3 chosen members are men:P(MMM) = (6/16) * (5/15) * (4/14)Simplifying the fractions:P(MMM) = (3/8) * (1/3) * (2/7)
Multiplying out and canceling common terms:P(MMM) = (1/8) * (2/7) = 2/56 = 1/28

GridLock · Answer

No. of males = 6
No. of females > 6
Let the number of females = f

Probability of selecting 2 people from same sex = 1/2
= Probability of selecting 2 men + Probability of selecting 2 women
=  *  +  * 
=> (f^2 - f + 30) / (f^2 + 11f + 30) = 1/2
=> f^2 - 13f + 30 = 0
=> (f-10)(f-3) = 0
=> f = 10 or 3

Since we know that f>6, then f must be 10

With f=10, the probability of selecting 3 men = 6/16 * 5/15 * 4/14 = 1/28

kevincan · Answer

We can also reason that the probability that the two chosen are of opposite sex is 1/2

GridLock · Answer

Yes, this is arguably an easier way to obtain the equation.

Probability of selecting members from opposite gender 
=  /  *    (for when the male is selected first) +  /  *  /  (for when female is selected first) = 1/2
=> 24f = f^2 + f + 30
=> f^2 - 13f + 30 = 0

from this point onwards, the approach remains the same.

findingmyself · Answer

My essential doubt here is that: We are said that three members are chosen at random. Hence the possible combinations are 8 (2*2*2)

Out of that, same sex in first two combinations are:
MMM, MMF, FFM, FFF (The probability that the first two chosen are of the same sex), here it is not said anything about third, so third can be either M or F

Now MMM= 6*5*4/(N*N-1*N-2)
MMF= 6*5*N-6/(N*N-1*N-2)
FFM= N-6*N-7*6/(N*N-1*N-2)
FFF= N-6*N-7*N-8/(N*N-1*N-2)

Sum of these 4 is equal to 1/2

When Im proceeding from here, it is becoming a very complex equation. How is this solvable?
The six male board members of a certain company are outnumbered by the female board members. If three board members are chosen at random, the probability that the first two chosen are of the same sex is 1/2. What is the probability that all 3 chosen members are men?

findingmyself · Answer

brunel  Is this type of questions common in GMAT? Can you check this please and give your POV,

kevincan · Answer

If the probability that the first two chosen members are of the same sex is  \frac{1}{2} , then the probability that they are of opposite sex is also  \frac{1}{2} .

Let the number of female board members be  w . Since there are 6 male members, the total number of board members is  w+6 .

The probability that the first two chosen members are of opposite sex is:

\frac{6}{w+6}\frac{w}{w+5}+\frac{w}{w+6}\frac{6}{w+5} =\frac{1}{2}

Thus,

\frac{12w}{(w+6)(w+5)}=\frac{1}{2}

24w=(w+6)(w+5)

w^2-13w+30=0

(w-3)(w-10)=0

So  w=3  or  w=10 . Since the female members outnumber the 6 male members,  w=10 .

Therefore, there are 16 board members, and the probability that all 3 chosen members are men is:

\frac{6}{16}\frac{5}{15}\frac{4}{14}=\boxed{\frac{1}{28}}

The six male board members of a certain company are outnumbe

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