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A small company employs 3 men and 5 women. If a team of 4 em

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by sudhir3127 » Tue Jul 15, 2008 10:18 am
is the answer D
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by pranavc » Tue Jul 15, 2008 12:55 pm
The answer should be D. A confirmation of this would be highly appreciated. Thanks in advance.
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by mim3 » Tue Jul 15, 2008 1:01 pm
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
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by sudhir3127 » Tue Jul 15, 2008 11:18 pm
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.
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by Sprite_TM » Tue May 19, 2009 7:36 pm
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.
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by barnum » Wed May 20, 2009 10:03 am
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.
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by raleigh » Wed May 20, 2009 10:55 am
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.
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by eski » Mon Oct 01, 2012 10:30 pm
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 ???
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by LalaB » Mon Oct 01, 2012 11:24 pm
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
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by anuprajan5 » Tue Oct 02, 2012 1:14 am
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
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by eski » Tue Oct 02, 2012 6:04 am
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?

:)
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by Brent@GMATPrepNow » Tue Oct 02, 2012 6:50 am
airan wrote: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
Solution via counting methods:

P(exactly 2 women) = [# of teams with exactly 2 women] / [total # of teams possible]

# of teams with exactly 2 women
Take the task of selecting 2 women and 2 men and break it into stages.
Stage 1: Select 2 women for the team. There are 5 women to choose from, so this can be accomplished in 5C2 ways.
Stage 2: Select 2 men for the team. There are 3 men to choose from, so this can be accomplished in 3C2 ways.

By the Fundamental Counting Principle (FCP), the total number of teams with exactly 2 women = (5C2)(3C2) = (10)(3) = 30

# of teams possible
There are 8 people altogether and we must choose 4 of them.
This can be accomplished in 8C4 ways, which equals 70 ways


P(exactly 2 women) = [30] / [70]
= 3/7 = D

Aside: To learn how to calculate combinations such as 5C2 in your head, you can watch the following free video: https://www.gmatprepnow.com/module/gmat-counting?id=789

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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