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If 2 different representatives are to be selected at random [#permalink]
08 Jan 2010, 13:18

If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

Re: If 2 different representatives [#permalink]
08 Jan 2010, 14:25

3

This post received KUDOS

Expert's post

sagarsabnis wrote:

If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

What is the probability of choosing 2 women out of 10 people \frac{w}{10}*\frac{w-1}{9} and this should be >1/2. So we have \frac{w}{10}*\frac{w-1}{9}>\frac{1}{2} --> w(w-1)>45 this is true only when w>7. (w # of women <=10)

So basically question asks is w>7?

(1) w>5 not sufficient.

(2) \frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<9 --> w>6, not sufficient

Re: If 2 different representatives [#permalink]
08 Jan 2010, 15:49

1

This post received KUDOS

Expert's post

sagarsabnis wrote:

eehhhh!!! I didnt read the question properly and I was using 10C2 for calculation

You can use C, for solving as well:

C^2_w # of selections of 2 women out of w employees;

C^2_{10} total # of selections of 2 representatives out of 10 employees.

Q is \frac{C^2_w}{C^2_{10}}>\frac{1}{2} --> \frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2} --> --> w(w-1)>45 --> w>7?

(1) w>5, not sufficient.

(2) C^2_{(10-w)} # of selections of 2 men out of 10-w=m employees --> \frac{C^2_{(10-w)}}{C^2_{10}}<\frac{1}{10} --> \frac{\frac{(10-w)(10-w-1)}{2}}{45}<\frac{1}{10} --> (10-w)(9-w)<9 --> w>6, not sufficient

Re: If 2 different representatives [#permalink]
23 Jun 2010, 12:40

Bunuel wrote:

sagarsabnis wrote:

(2) \frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<9 --> w>6, not sufficient

Answer E.

Hi Bunuel,

How did you get w>6. I tried solving it by moving 9 on left side and expanding (w-10) (w-9)-9<0 w^2-19w+81<0 If we solve this quadratic we don't get a whole number for w.

Could you please elaborate? Your approach is very good as always.

Re: If 2 different representatives [#permalink]
23 Jun 2010, 12:54

2

This post received KUDOS

Expert's post

sevenplus wrote:

Bunuel wrote:

sagarsabnis wrote:

(2) \frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<9 --> w>6, not sufficient

Answer E.

Hi Bunuel,

How did you get w>6. I tried solving it by moving 9 on left side and expanding (w-10) (w-9)-9<0 w^2-19w+81<0 If we solve this quadratic we don't get a whole number for w.

Could you please elaborate? Your approach is very good as always.

As you correctly noted w must be an integer (as w represents # of women). Now, substituting: if w=6, then (10-w)(9-w)=(10-6)(9-6)=12>9, but if w>6, for instance 7, then (10-w)(9-w)=(10-7)(9-7)=6<9. So, w>6.

Re: If 2 different representatives [#permalink]
11 Jul 2010, 04:53

Bunuel wrote:

sagarsabnis wrote:

If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

What is the probability of choosing 2 women out of 10 people \frac{w}{10}*\frac{w-1}{9} and this should be >1/2. So we have \frac{w}{10}*\frac{w-1}{9}>\frac{1}{2} --> w(w-1)>45 this is true only when w>7. (w # of women <=10)

So basically question asks is w>7?

(1) w>5 not sufficient.

(2) \frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<9 --> w>6, not sufficient

(1)+(2) w>5, w>6 not sufficient

Answer E.

hi, since from basic data we find that w>7 and 2. statemant state that at least 2 men are there out of 10 ........ it is asking that . probability of choosing 2 men is 1/10 . ie. atleast 2 men are there. .... so, now combining w>7 and two men the only option left is w=8. hence ans is B

Re: If 2 different representatives [#permalink]
11 Jul 2010, 06:44

1

This post received KUDOS

Expert's post

wrldcabhishek wrote:

Bunuel wrote:

sagarsabnis wrote:

If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

What is the probability of choosing 2 women out of 10 people \frac{w}{10}*\frac{w-1}{9} and this should be >1/2. So we have \frac{w}{10}*\frac{w-1}{9}>\frac{1}{2} --> w(w-1)>45 this is true only when w>7. (w # of women <=10)

So basically question asks is w>7?

(1) w>5 not sufficient.

(2) \frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<9 --> w>6, not sufficient

(1)+(2) w>5, w>6 not sufficient

Answer E.

hi, since from basic data we find that w>7 and 2. statemant state that at least 2 men are there out of 10 ........ it is asking that . probability of choosing 2 men is 1/10 . ie. atleast 2 men are there. .... so, now combining w>7 and two men the only option left is w=8. hence ans is B

OA for this question is E. The problem with your solution is that we are not given that w>7, in fact we are asked to find whether this is true. _________________

Re: If 2 different representatives [#permalink]
12 Jul 2010, 10:47

Bunuel wrote:

sagarsabnis wrote:

If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 1/10

What is the probability of choosing 2 women out of 10 people \frac{w}{10}*\frac{w-1}{9} and this should be >1/2. So we have \frac{w}{10}*\frac{w-1}{9}>\frac{1}{2} --> w(w-1)>45 this is true only when w>7. (w # of women <=10)

So basically question asks is w>7?

(1) w>5 not sufficient.

(2) \frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<9 --> w>6, not sufficient

(1)+(2) w>5, w>6 not sufficient

Answer E.

you rock buddy

_________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

Re: If 2 different representatives are to be selected at random [#permalink]
02 Jul 2014, 09:34

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