Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one.
A small company employs 3 men and 5 women. If a team of 4
[#permalink]
Show Tags
22 Nov 2007, 18:46
13
57
00:00
A
B
C
D
E
Difficulty:
45% (medium)
Question Stats:
66% (01:15) correct 34% (01:23) wrong based on 1580 sessions
HideShow timer Statistics
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?
SECOND METHOD: Probability of two women -> 5/8 * 4/7.
probability of two men -> 3/6 * 2/5.
Probability: (5/8 * 4/7) * (3/6 * 2/5) = 1/14.
I know something is wrong with the second method but i can't really figure out why its flawed.
Any pointers ?? thanks
The problem here is that you are arranging the people. When you select a woman out of 8 as 5/8, you are saying that you are picking a woman first. You are arranging them in this way: WWMM Now, if you want to un-arrange them, multiply it by the total number of arrangements i.e. 4!/(2!*2!) (because there are 2 men and 2 women so you divide by 2!s to get the total number of arrangements)
When you do that, you get 1/14 * 4!/(2!*2!) = 3/7
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Schools: Haas (WL), Kellogg (matricultating), Stanford (R2, ding), Columbia (ding)
WE 1: 3 years hotel industry sales and marketing France
WE 2: 3 years financial industry marketing UK
A small company employs 3 men and 5 women. If a team of 4
[#permalink]
Show Tags
23 May 2010, 11:11
1
7
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
I find the explanation provided in MGMAT very complicated and would like to know what techniques other people would recommend to solve this problem in 2 mins.
_________________
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
I find the explanation provided in MGMAT very complicated and would like to know what techniques other people would recommend to solve this problem in 2 mins.
Will post OA later.
P=Favorable outcomes / Total # of outcomes
Favorable outcomes: 2 women and 2 men = \(C^2_5*C^2_3=30\); Total # of outcomes: 4 people out of 5+3=8 = \(C^4_8=70\);
I was working on a practice test when i saw this question. Now, the problem is; i have solved it in two ways, one way(using combinations) leads to the correct answer, the other gives me something different. i would appreciate if you could point out what im doing wrong here.
Questn 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
first method: 5C2 * 3C2 -> it gives combination of exactly 2 women and 2 men. 8C4 -> gives total possibilities of 4 people from 5 women and 3 men.
Probability = 5C2*3C2 / 8C4 = 3/7
SECOND METHOD: Probability of two women -> 5/8 * 4/7.
probability of two men -> 3/6 * 2/5.
Probability: (5/8 * 4/7) * (3/6 * 2/5) = 1/14.
I know something is wrong with the second method but i can't really figure out why its flawed.
SECOND METHOD: Probability of two women -> 5/8 * 4/7.
Quote:
The problem here is that you are arranging the people. When you select a woman out of 8 as 5/8, you are saying that you are picking a woman first. You are arranging them in this way: WWMM Now, if you want to un-arrange them, multiply it by the total number of arrangements i.e. 4!/(2!*2!) (because there are 2 men and 2 women so you divide by 2!s to get the total number of arrangements)
When you do that, you get 1/14 * 4!/(2!*2!) = 3/7
Hello karishma,
From what i understand, you have added all probabilities like so: 1) WWMM 2) WMMW 3) WMWM 4) MWMW 5) MMWW 6) MWWM
...such that you do not repeat arrangements that are the same, such as W1W2 and W2W1. ... right ? Thanks
Yes. What you have to do is select a group. Hence you do not have to arrange them. In your second method, you got 1/14 which is the probability of getting wwmm. Since there are other such arrangements too which are all acceptable to us e.g. Wmwm since we just need a group of 2 men and 2 women irrespective of their arrangement, we multiply 1/14 by 6 (since 6 such arrangements are possible as shown by you)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
Show Tags
28 Nov 2014, 01:09
Bunuel,i did not understand why we took two men too?I did it like Favorable outcomes: selecting 2 women C^2_5 and Total # of outcomes: 4 people out of 5+3=8 = C^4_8=70;
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
Show Tags
28 Nov 2014, 04:11
Ralphcuisak wrote:
Bunuel,i did not understand why we took two men too?I did it like Favorable outcomes: selecting 2 women C^2_5 and Total # of outcomes: 4 people out of 5+3=8 = C^4_8=70;
The team should have 4 employees out of which 2 are women, thus the remaining 2 will be men.
_________________
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
Show Tags
14 Dec 2017, 09:13
alimad 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
We need to determine the probability that when 4 employees are selected from 3 men and 5 women, 2 women and 2 men will be selected.
Let’s determine the number of ways to select the 2 women and 2 men.
Number of ways to select 2 women:
5C2 = (5 x 4)/2! = 10
Number of ways to select 2 men:
3C2 = (3 x 2)/2! = 3
Thus, 2 men and 2 women can be selected in 10 x 3 = 30 ways.
Finally, we can determine the number of ways to select 4 people from 8.
8C4 = 8! / [4! x (8-4)!] = (8x 7 x 6 x 5)/4! = (8 x 7 x 6 x 5)/(4 x 3 x 2) = 7 x 2 x 5 = 70
Thus, the probability is 30/70 = 3/7.
Answer: D
_________________
Jeffery Miller Head of GMAT Instruction
GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
Show Tags
06 Jan 2018, 08:35
x2suresh wrote:
\(P=\frac{3C2*5C2}{8C4}\)
Could someone please explain the way how to solve this fraction mathematically? What does C2 and C4 stand for? What calculations are derived from this?
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
Show Tags
06 Jan 2018, 09:17
1
Gregsterh wrote:
x2suresh wrote:
\(P=\frac{3C2*5C2}{8C4}\)
Could someone please explain the way how to solve this fraction mathematically? What does C2 and C4 stand for? What calculations are derived from this?
Thanks
C stands for combinations: \(xCy=\frac{x!}{y!(x-y)!}=21\).
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
Show Tags
12 Jan 2018, 13:01
Hi All,
You did the first 'step' of this question correctly. The number of groups of 2 men IS 3 and the number of groups of 2 women IS 10. Now, you have to do the rest of the math in a really specific way:
Since each "subgroup" of men can be paired with each "subgroup" of women, there are (3)(10) = 30 possible groups of four that include exactly 2 women. Since this is a probability question, we now have to determine the total number of groups of four that are possible...
That is 8C4 = (8)(7)(6)(5) / (4)(3)(2)(1) = (2)(7)(1)(5) = 70 possible groups of 4
close
66% (01:15) correct 
