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?
1/14
1/7
2/7
3/7
1/2
I would appreciate if someone can try it without using counting method. Thanks!
GMAT Forum
9 postsPage 1 of 1
- ahistegt
- MICHAEL_SHAUNN
counting not allowed in probability....
HI,
HONESTLY SPEAKING,PROBABILITY HAS GOT A GREAT DEAL OF CONNECTION WITH PERMUTATION AND COMBINATION.
HERE,I SOLVE IT USING A BIT OF COMBINATION.
NOT,OUT OF 3M AND 5W,A COMMITTEE OF 4 MEMBERS CAN BE SELECTED IN 8C4=70 WAYS.(no condition appplied)
AND THE NUMBER OF COMMITTEES WITH EXACTLY 2 WOMEN IS 5C2*3C2=30 WAYS.(first impose the condition that is select the women and then the rest members)
THE ANSWER SHOULD BE 30/70=3/7.
THANK U AND GOOD LUCK.
HONESTLY SPEAKING,PROBABILITY HAS GOT A GREAT DEAL OF CONNECTION WITH PERMUTATION AND COMBINATION.
HERE,I SOLVE IT USING A BIT OF COMBINATION.
NOT,OUT OF 3M AND 5W,A COMMITTEE OF 4 MEMBERS CAN BE SELECTED IN 8C4=70 WAYS.(no condition appplied)
AND THE NUMBER OF COMMITTEES WITH EXACTLY 2 WOMEN IS 5C2*3C2=30 WAYS.(first impose the condition that is select the women and then the rest members)
THE ANSWER SHOULD BE 30/70=3/7.
THANK U AND GOOD LUCK.
- michael_shaunn
- JonathanSchneider
- ManhattanGMAT Staff
- Posts: 370
- Joined: Sun Oct 26, 2008 3:40 pm
Actually, ahestegt, you've got a great approach there.
I'll take it that you mean to use the "slot method" as follows:
The probability of getting a man on your first pick: 3/8
The probability of getting another man on your second pick: 2/7 (because we have fewer people now)
The probability of getting a woman on your third pick: 5/6 (because there are only six total people remaining)
The probability of getting another woman on your fourth pick: 4/5
You are right to multiply these together, as we need to make all four picks. The resulting product is indeed 1/14.
So why is that not the answer?
Well, notice that you have picked them in order, as if order mattered. In other words, one out of every 14 combinations will have two men first, then two women second. HOWEVER, what if the men and women were not chosen exactly in that order? What if it were WWMM, WMWM, MWMW, MWWM, or WMMW instead? (These are the only five other possibile arrangements.) Well, you'd have to do the same math again, getting to 1/14 for each of those arrangements. And then you'd have to ADD those 6 fractions of 1/14 together (because we would want either the first arrangement OR the second arrangement OR the third arrangement...). You could thus multiply your 1/14 by 6 to get 3/7.
However, do you really still think that this is the simplest approach? At the very least, save yourself some time by noticing that we can arrange the two men and the two women 6 different ways, by showing 4!/(2!2!) = 6.
I'll take it that you mean to use the "slot method" as follows:
The probability of getting a man on your first pick: 3/8
The probability of getting another man on your second pick: 2/7 (because we have fewer people now)
The probability of getting a woman on your third pick: 5/6 (because there are only six total people remaining)
The probability of getting another woman on your fourth pick: 4/5
You are right to multiply these together, as we need to make all four picks. The resulting product is indeed 1/14.
So why is that not the answer?
Well, notice that you have picked them in order, as if order mattered. In other words, one out of every 14 combinations will have two men first, then two women second. HOWEVER, what if the men and women were not chosen exactly in that order? What if it were WWMM, WMWM, MWMW, MWWM, or WMMW instead? (These are the only five other possibile arrangements.) Well, you'd have to do the same math again, getting to 1/14 for each of those arrangements. And then you'd have to ADD those 6 fractions of 1/14 together (because we would want either the first arrangement OR the second arrangement OR the third arrangement...). You could thus multiply your 1/14 by 6 to get 3/7.
However, do you really still think that this is the simplest approach? At the very least, save yourself some time by noticing that we can arrange the two men and the two women 6 different ways, by showing 4!/(2!2!) = 6.
- JonathanSchneider
- ManhattanGMAT Staff
- Posts: 477
- Joined: Wed Dec 12, 2007 5:40 am
- Location: Durham, NC
Re: CAT 3
: )
- epic_one75
- Course Students
- Posts: 1
- Joined: Sun Mar 02, 2008 12:14 pm
Re: CAT 3
Here's my question on this: why can this question not be solved by using the basic probability formula of desired outcomes / all possible outcomes ?
In this case, desired outcomes is two women, which I had as 5C2. Desired outcomes is 8C4, i.e. picking 4 people out of 8.
The final answer for me came out to 1/7
In this case, desired outcomes is two women, which I had as 5C2. Desired outcomes is 8C4, i.e. picking 4 people out of 8.
The final answer for me came out to 1/7
- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
Re: CAT 3
epic_one75 Wrote:In this case, desired outcomes is two women, which I had as 5C2. Desired outcomes is 8C4, i.e. picking 4 people out of 8.
this doesn't work because your team doesn't consist only of 2 women.
your team consists of two women and two men. since there are multiple ways in which to pick two men, you MUST include that in your calculations, or else you will be massively underestimating the number of "successful" selections.
you are correct that the number of ways to choose the women only is 5c2 (or (5x4)/(2!), if you use the slot method). this reduces to 10.
however, the number of ways to choose the MEN (whom you must also choose) is 3c2 (or, if you use the slot method, (3x2)/(2!). this reduces to 3.
therefore, it's not just 10/70, but rather (10 x 3) / 70, or 3/7.
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