Problem Solving

A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

a. 1/14
b. 1/7
c. 2/7
d. 3/7
e. 1/2

is the answer D

The answer should be D. A confirmation of this would be highly appreciated. Thanks in advance.

Agreed- D.

prob= #winning solutions/total # possible.

total # of possible combintions of 4 people chosen from 8:
8!/(4!)(4!)= 70

total # of winning solutions= # ways to get a team of 2 men and 2 women.

# ways to get 2 men: 3!/(2!)(1!)= 3
# ways to get 2 women: 5!/(3!)(2!)= 10

(3)(10)/70 = 30/70 = 3/7

Answer D .. here it goes.

probability of choosing 2 women out of 5 is 5 C2= 10
probability of choosing 2 men out of 3 is 3 C2 = 3

total probability of choosing 4 out 8 is 8 C4= 70

hence its (10*3)/70 = 3/7

hope it helps.

can someone explain why you divide 8! / 4! 4!. i don't understand the bottom part

edit: intially i thought like this

Answer = A (1/14).

Probability of selecting 1st woman: (5/8)
Prob of 2nd woman: (4/7)
Prob of 1st man: (3/6)
Prob of 2nd man: (2/5)

Probability of selecting exactly 2 women: (5/8)(4/7)(3/6)(2/5)=1/14

edit: i think i get it now, its just that the repeat of 4! confused me.

Sprite_TM wrote:can someone explain why you divide 8! / 4! 4!. i don't understand the bottom part

edit: intially i thought like this

Answer = A (1/14).

Probability of selecting 1st woman: (5/8)
Prob of 2nd woman: (4/7)
Prob of 1st man: (3/6)
Prob of 2nd man: (2/5)

Probability of selecting exactly 2 women: (5/8)(4/7)(3/6)(2/5)=1/14

edit: i think i get it now, its just that the repeat of 4! confused me.

The problem with this thought process is that it presumes a specific order to choosing the people (i.e. that you will definitely select in the order WWMM). However there are six different orders you could do

WWMM
WMWM
WMMW
MWWM
MWMW
MMWW

Each one would then calculate out to having a 1/14 chance (what you calculated earlier). Since there were 6 different ways you could do it you would multiply 1/14*6 to get the same 3/7 that was calculated earlier in this thread. So you could do it this way, it is just a bit more difficult and riskier if you don't realize you have to account for six different orderings.

There is only one case to get this result. That happens by choosing 2 women, and 2 men for the team. Order definitely doesn't matter so this is a combination problem. So 5 choose 2 for women, times 3 choose 2 for men, and there are a total of 8 choose 4 possibilities.

(5C2)(3C2)/(8C4) which simplifies to 3/7.

Well now I get a ans , can someone clarrify what Im doing wrong here.

Using slot method.

1st slot : select 1 women from 8 ppl out of 5 women = 5/8
2nd slot : selecting 2nd women = 4/7
3rd slot : selecting 1 man from remaining ppl out of 3 men = 3/6
4th slot : selecting 2nd man = 2/5

5/8*4/7*3/6*2/5 = 1/14 (infact one of the ans choice )

that im doing wrong here ???

eski wrote:Well now I get a ans , can someone clarrify what Im doing wrong here.

Using slot method.

1st slot : select 1 women from 8 ppl out of 5 women = 5/8
2nd slot : selecting 2nd women = 4/7
3rd slot : selecting 1 man from remaining ppl out of 3 men = 3/6
4th slot : selecting 2nd man = 2/5

5/8*4/7*3/6*2/5 = 1/14 (infact one of the ans choice )

that im doing wrong here ???

U found the Probability of WWMM only, but u should think of the rest variations.

So, u should 4!/2!*2!=6
(1/14)*6=3/7

u can also read here -https://www.beatthegmat.com/mgmat-probabilty-t87745.html

You need to find how many ways you can arrange exactly 2 women from 4 people.

4C2 = 6

To illustrate

WWMM
WMWM
WMMW
MMWW
MWWM
MWMW

6 ways to arrange the selection of 2 women.

So - 6*1/14 = 3/7

Regards
Anup

Great made my own formula :

1. Is selection is based on the order , select SLOT METHOD.

2. Any random picks (without any order) use combination method.


my neurons got wired for one more logic flow. any common place where i can find such deduction , any rule books?