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# A committee of 2 people is to be formed from a group of 8

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A committee of 2 people is to be formed from a group of 8  [#permalink]

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22 Jan 2011, 03:06
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A committee of 2 people is to be formed from a group of 8 people which includes some women and rest are men. If P is the probability that both the selected people are men, is P > 0.25?

(1) More than 40% of the employees are men.
(2) The probability that both the selected people will be women is more than 20%.
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Re: A committee of 2 people, is P > 0.25?  [#permalink]

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22 Jan 2011, 03:32
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MichelleSavina wrote:
Q) A committee of 2 people is to be formed from a group of 8 people which includes some women and rest are men. If P is the probability that both the selected people are men, is P > 0.25?
(1) More than 40% of the employees are men.
(2) The probability that both the selected people will be women is more than 20%.

Given: $$m+w=8$$. Question: is $$\frac{m}{8}*\frac{m-1}{7}>\frac{1}{4}$$? --> is $$m(m-1)>14$$ --> is $$m\geq{5}$$? So the question basically asks whether the # of men in 8 people is more than or equal to 5 (5, 6, or 7).

(1) More than 40% of the employees are men --> $$m>0.4*8$$ --> $$m>3.2$$: there may be 4 men and in this case the answer to the question will be NO or there may be more than 4 men (5, 6, ...) and in this case the answer to the question will be YES. Not sufficient.

(2) The probability that both the selected people will be women is more than 20% -> $$\frac{w}{8}*\frac{w-1}{7}>\frac{1}{5}$$ --> $$w(w-1)>11.2$$ --> is $$w\geq{4}$$: there are more than or equal to 4 women in 8 people: 4, 5, 6, ... hence there are less than or equal to 4 men: 4, 3, ... So answer to the question whether there are more than or equal to 5 men is NO. Sufficient.

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Re: A committee of 2 people is to be formed  [#permalink]

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19 May 2012, 09:39
gmihir wrote:
A committee of 2 people is to be formed from a group of 8 people which includes some women and rest are men.If P is the probability that both the selected people are men, is P > 0.25?
(1) More than 40% of the employees are men.
(2) The probability that both the selected people will be women is more than 20%.

This is my first answer here... so I dont know how to use maths symbols here...

Probability that both men are selected is MC2/8C2. We have to find whether this is greater than 1/4.
If you solve : MC2/8C2 > 1/4, it gives you M^2 - M - 14 >0 which is true for M > 4.

So the first option is ruled out because 40% of 8 is 3.2, which means M can either 4 or greater than 4.

2. If you solve the second option, WC2/8C2 > 1/2, it gives W^2 - W- 11 > 0, which is true for W >= 4. Since total people are 8, M <= 4.

So, the answer is B.

Hope its clear.
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Re: A committee of 2 people is to be formed from a group of 8  [#permalink]

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20 May 2012, 10:48
(2) The probability that both the selected people will be women is more than 20% -> \frac{w}{8}*\frac{w-1}{7}>\frac{1}{5} --> w(w-1)>11.2 --> is w\geq{4}: there are more than or equal to 4 women in 8 people: 4, 5, 6, ... hence there are less than or equal to 4 men: 4, 3, ... So answer to the question whether there are more than or equal to 5 men is NO. Sufficient.

can you explain what this equaiton means? W(W-1) >11.2 How do we know its SUFF by this equation?
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Re: A committee of 2 people is to be formed from a group of 8  [#permalink]

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20 May 2012, 11:27
gmihir wrote:
A committee of 2 people is to be formed from a group of 8 people which includes some women and rest are men. If P is the probability that both the selected people are men, is P > 0.25?

(1) More than 40% of the employees are men.
(2) The probability that both the selected people will be women is more than 20%.

Merging similar topics. Please refer to the solution above.

dchow23 wrote:
(2) The probability that both the selected people will be women is more than 20% -> \frac{w}{8}*\frac{w-1}{7}>\frac{1}{5} --> w(w-1)>11.2 --> is w\geq{4}: there are more than or equal to 4 women in 8 people: 4, 5, 6, ... hence there are less than or equal to 4 men: 4, 3, ... So answer to the question whether there are more than or equal to 5 men is NO. Sufficient.

can you explain what this equaiton means? W(W-1) >11.2 How do we know its SUFF by this equation?

The question asks whether the # of men in 8 people is more than or equal to 5 (5, 6, or 7). From (2) we have that $$w(w-1)>11.2$$. Now, if the # of women is 3 then 3(3-1)=6<11.2, so w>3, which means that there are more than or equal to 4 women in 8 people: 4, 5, 6, ... hence there are less than or equal to 4 men: 8-4=4, 8-5=3, ... So, the answer to the question whether there are more than or equal to 5 men is NO.

Hope it's clear.
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Re: A committee of 2 people is to be formed from a group of 8  [#permalink]

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15 Jun 2012, 03:07
Good question and great explanation... thanks
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Re: A committee of 2 people is to be formed from a group of 8  [#permalink]

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01 Apr 2015, 20:38
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1
Hi All,

There are a couple of ways to approach this question, depending on how comfortable you are with the math concepts involved. There's a great 'brute force' element to the question that you can take advantage of (as long as you write everything down and label your work).

We're told that we have a group of 8 people (some men and some women). We're told to randomly select 2 of the 8 people and we're asked if the probability that they're BOTH men is greater than .25 This is a YES/NO question.

Since there are only 8 people, we can have from 0 - 8 men in the group. We can do a bit of work up-front so that we can move faster through the two Facts later. We want to figure out the point at which the probability will be > .25

IF....
There are 4 men and 4 women
Probability of selecting 2 men is (4/8)(3/7) = 3/14
Since 3/12 = .25, 3/14 must be LESS than .25 (since the denominator of that fraction is bigger than 12)

IF.....
There are 5 men and 3 women
Probability of selecting 2 men is (5/8)(4/7) = 5/14
Since 5/15 = .333333, 5/14 must be BIGGER than .3333 (since the denominator of that fraction is smaller than 15)

So, with 4 OR FEWER men, the answer to the question is NO
With 5 OR MORE men, the answer to the question is YES

We now know that the question is really asking if the number of men is 5 OR GREATER. Again, this is a YES/NO question.

Fact 1: MORE than 40% of the employees are men.

(.4)(8) = 3.2, BUT since we CANNOT have a "fraction" of a man, we must have 4 OR MORE men....

IF...we have 4 men, then the answer to the question is NO
IF...we have 5 men, then the answer to the question is YES
Fact 1 is INSUFFICIENT

(2) The probability that both the selected people will be women is more than 20%.

For this Fact, we can use the calculations that we did at the beginning (with a little extra work).

IF....
There are 4 women and 4 men
Probability of selecting 2 women is (4/8)(3/7) = 3/14
Since 3/15 = .2, 3/14 must be GREATER than .2 (since the denominator of that fraction is less than 15)

This tells us that we have 4 OR MORE women.

IF... we have 4 women, then the answer to the question is NO
IF... we have MORE women, we have even FEWER men and the answer to the question is still NO.
Fact 2 is SUFFICIENT

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Re: A committee of 2 people is to be formed from a group of 8  [#permalink]

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18 May 2015, 04:10
1
Bunuel wrote:
MichelleSavina wrote:
Q) A committee of 2 people is to be formed from a group of 8 people which includes some women and rest are men. If P is the probability that both the selected people are men, is P > 0.25?
(1) More than 40% of the employees are men.
(2) The probability that both the selected people will be women is more than 20%.

Given: $$m+w=8$$. Question: is $$\frac{m}{8}*\frac{m-1}{7}>\frac{1}{4}$$? --> is $$m(m-1)>14$$ --> is $$m\geq{5}$$? So the question basically asks whether the # of men in 8 people is more than or equal to 5 (5, 6, or 7).

(1) More than 40% of the employees are men --> $$m>0.4*8$$ --> $$m>3.2$$: there may be 4 men and in this case the answer to the question will be NO or there may be more than 4 men (5, 6, ...) and in this case the answer to the question will be YES. Not sufficient.

(2) The probability that both the selected people will be women is more than 20% -> $$\frac{w}{8}*\frac{w-1}{7}>\frac{1}{5}$$ --> $$w(w-1)>11.2$$ --> is $$w\geq{4}$$: there are more than or equal to 4 women in 8 people: 4, 5, 6, ... hence there are less than or equal to 4 men: 4, 3, ... So answer to the question whether there are more than or equal to 5 men is NO. Sufficient.

Hi Bunuel

Why is ,the total number of ways in which 2 people can be selected , 8*7 and not 8C2?
that would mean we are taking order into consideration!
For ex : out of 4 people A B C D

ways of choosing 2 people - to form a committee - is 4C2 = 6
AB AC AD BC BD CD

and not 4*3 => which is 12 and will include the reverse as well since there isn't any position or order mentioned here while forming the committee.
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Re: A committee of 2 people is to be formed from a group of 8  [#permalink]

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18 May 2015, 04:48
2
Anu26 wrote:
Bunuel wrote:
MichelleSavina wrote:
Q) A committee of 2 people is to be formed from a group of 8 people which includes some women and rest are men. If P is the probability that both the selected people are men, is P > 0.25?
(1) More than 40% of the employees are men.
(2) The probability that both the selected people will be women is more than 20%.

Given: $$m+w=8$$. Question: is $$\frac{m}{8}*\frac{m-1}{7}>\frac{1}{4}$$? --> is $$m(m-1)>14$$ --> is $$m\geq{5}$$? So the question basically asks whether the # of men in 8 people is more than or equal to 5 (5, 6, or 7).

(1) More than 40% of the employees are men --> $$m>0.4*8$$ --> $$m>3.2$$: there may be 4 men and in this case the answer to the question will be NO or there may be more than 4 men (5, 6, ...) and in this case the answer to the question will be YES. Not sufficient.

(2) The probability that both the selected people will be women is more than 20% -> $$\frac{w}{8}*\frac{w-1}{7}>\frac{1}{5}$$ --> $$w(w-1)>11.2$$ --> is $$w\geq{4}$$: there are more than or equal to 4 women in 8 people: 4, 5, 6, ... hence there are less than or equal to 4 men: 4, 3, ... So answer to the question whether there are more than or equal to 5 men is NO. Sufficient.

Hi Bunuel

Why is ,the total number of ways in which 2 people can be selected , 8*7 and not 8C2?
that would mean we are taking order into consideration!
For ex : out of 4 people A B C D

ways of choosing 2 people - to form a committee - is 4C2 = 6
AB AC AD BC BD CD

and not 4*3 => which is 12 and will include the reverse as well since there isn't any position or order mentioned here while forming the committee.

Dear Anu26

You're absolutely right.

The probability of selecting 2 men in the group = (No. of ways in which 2 men can be selected)/(No. of ways in which 2 people can be selected)

=mC2/8C2

= $$\frac{m(m-1)}{8*7}$$ (Since both mC2 and 8C2 contain 2!, it cancels out to give us this expression)

Hope this helped!

Regards

Japinder
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A committee of 2 people is to be formed from a group of 8  [#permalink]

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18 May 2015, 05:08
1
Ah ok right! Thanks. I didn't see m(m-1) in the numerator in the solution. Just jumped to why is it 8*7

So then with numbers starting 4 - considering statement 1) - with men=4 -> P = 4C2/8C2 = 6/28 , P <0.25

and then with 5 men -> P = 5C2/8C2 = 10/28 ,P>0.25 so that tells me how it is insufficient!

Thanks much. +1 to you!
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