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
A small company employs 3 men and 5 women. If a team of 4 em
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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
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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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.
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.
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 doSprite_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.
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.
(5C2)(3C2)/(8C4) which simplifies to 3/7.
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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 ???
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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U found the Probability of WWMM only, but u should think of the rest variations.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 ???
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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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
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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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?
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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Solution via counting methods: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
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