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Probability Question

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by chrisjim5 » Mon Oct 18, 2010 6:48 am
Here is a question:

The employees of a certain company include exactly 4 men and 4 women. If 5 employees are to be selected at random, what is the probability that all of the women are selected?

Please let me know how to solve this ...

Thanks ...
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by kmittal82 » Mon Oct 18, 2010 7:01 am
Total employeed = 8
Number to be selected = 5

Total ways to select 5 employees out of 8 = 8C5

#ways of selecting all women = #ways of selecting 1 man * #ways of selecting 4 women = 4C1x4C4 = 4x1 = 4

Thus, probability = 4/8C5 = 4/56 = 1/14

OA?
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by GMATGuruNY » Mon Oct 18, 2010 7:15 am
chrisjim5 wrote:Here is a question:

The employees of a certain company include exactly 4 men and 4 women. If 5 employees are to be selected at random, what is the probability that all of the women are selected?

Please let me know how to solve this ...

Thanks ...
Another approach:

P(WWWWM) = 4/8 * 3/7 * 2/6 * 1/5 * 4/4 = 1/70.

Since M can appear in any of the 5 positions, we multiply by 5:

5*(1/70) = 1/14.
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by chrisjim5 » Mon Oct 18, 2010 7:20 am
Thats right ...

Thanks for explaining ...
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by shovan85 » Mon Oct 18, 2010 7:23 am
GMATGuruNY wrote:
chrisjim5 wrote:Here is a question:

The employees of a certain company include exactly 4 men and 4 women. If 5 employees are to be selected at random, what is the probability that all of the women are selected?

Please let me know how to solve this ...

Thanks ...
Another approach:

P(WWWWM) = 4/8 * 3/7 * 2/6 * 1/5 * 4/4 = 1/70.

Since M can appear in any of the 5 positions, we multiply by 5:

5*(1/70) = 1/14.
Why the following method is getting wrong Please clarify?

WWWWM, WWWMW, WWMWW, WMWWW, MWWWW = 5 possibilities with 4 W and 1 M

Total possibilities = 8C5 = 56

Why not 5/56 ?
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by Geva@EconomistGMAT » Mon Oct 18, 2010 7:33 am
shovan85 wrote:
GMATGuruNY wrote:
chrisjim5 wrote:Here is a question:

The employees of a certain company include exactly 4 men and 4 women. If 5 employees are to be selected at random, what is the probability that all of the women are selected?

Please let me know how to solve this ...

Thanks ...
Another approach:

P(WWWWM) = 4/8 * 3/7 * 2/6 * 1/5 * 4/4 = 1/70.

Since M can appear in any of the 5 positions, we multiply by 5:

5*(1/70) = 1/14.
Why the following method is getting wrong Please clarify?

WWWWM, WWWMW, WWMWW, WMWWW, MWWWW = 5 possibilities with 4 W and 1 M

Total possibilities = 8C5 = 56

Why not 5/56 ?
The two are just incompatible: 8C5 is the number of ways of choosing 5 out of 8, order doesn't matter plays around with combinations of all 4 men), while the 5 options actually assumes a single man and order matters. Whatever, you do, be consistent: either order matters in both top and bottom, or it doesn't.
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by GMATGuruNY » Mon Oct 18, 2010 7:35 am
shovan85 wrote:
GMATGuruNY wrote:
chrisjim5 wrote:Here is a question:

The employees of a certain company include exactly 4 men and 4 women. If 5 employees are to be selected at random, what is the probability that all of the women are selected?

Please let me know how to solve this ...

Thanks ...
Another approach:

P(WWWWM) = 4/8 * 3/7 * 2/6 * 1/5 * 4/4 = 1/70.

Since M can appear in any of the 5 positions, we multiply by 5:

5*(1/70) = 1/14.
Why the following method is getting wrong Please clarify?

WWWWM, WWWMW, WWMWW, WMWWW, MWWWW = 5 possibilities with 4 W and 1 M

Total possibilities = 8C5 = 56

Why not 5/56 ?
You're mixing up methods.

8C5 counts the number ways to combine 5 out of 8 choices.

WWWWM, WWWMW, WWMWW, WMWWW, MWWWW = 5 = the number of ways to arrange the 5 letters WWWWM. (We could also determine this by saying the number of ways to arrange WWWWM = 5!/4! = 5.)

You can't put arrangements/combinations.

Does this help?
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by abhishekg21 » Mon Oct 18, 2010 7:42 am
you are taking into account the order or arrangement.Where as in commitee it does not matter. wwwwm is same as mwwww.
so remove the arrangment from your answer and you will correct figure of 1/14
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by shovan85 » Mon Oct 18, 2010 7:49 am
Geva Stern wrote: The two are just incompatible: 8C5 is the number of ways of choosing 5 out of 8, order doesn't matter plays around with combinations of all 4 men), while the 5 options actually assumes a single man and order matters. Whatever, you do, be consistent: either order matters in both top and bottom, or it doesn't.
Finally I guess I got it...

5 possibilities if I take as WWWWM, WWWMW, WWMWW, WMWWW, MWWWW then order matters. Using Combinatorics I got 1/14 but the picture was not getting clear in my mind as I had taken W1,W2...

So I was taking the upper part with respect to Arrangement but was considering the formula of Combinatorics.

Using Permutations it will be 5*(4P4 * 4P1)/8P5 = 1/14.

Not a good idea but I have a clear picture now ;)

Thanks Geva Stern I was not able to see this
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by shovan85 » Mon Oct 18, 2010 7:51 am
GMATGuruNY wrote:You're mixing up methods.

8C5 counts the number ways to combine 5 out of 8 choices.

WWWWM, WWWMW, WWMWW, WMWWW, MWWWW = 5 = the number of ways to arrange the 5 letters WWWWM. (We could also determine this by saying the number of ways to arrange WWWWM = 5!/4! = 5.)

You can't put arrangements/combinations.

Does this help?
Thanks it helps a lot Never mind it was a stupid Question after all
and no more Maths for me today ;)
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