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# If 2 different representatives are to be selected at random

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Joined: 17 Jul 2014
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Re: If 2 different representatives are to be selected at random  [#permalink]

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09 Feb 2016, 18:26
oh damn..got myself screwed hard time on this one..and all because of a small mistake..
I took:
1) 6W so-> 6/10 * 5/9 = 2/3 but it should be 2/6 or 1/3 which is not sufficient.
2) tested few possibilities:
M=1 - works -> W=9 - yes
M=2 - 1/5*1/9 = 1/45 - works, W=8 - works
M=3 -> 3/10*2/9 = 1/15 - works, W=7 - works
M=4 -> 2/5*1/3 = 2/15 - works, W=6 -> again same mistake as in 1.
M=5 -> not work - so considered 2 sufficient.

because of a small calculation mistake, I got it wrong..god I hate myself....
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Joined: 10 Apr 2015
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Re: If 2 different representatives are to be selected at random  [#permalink]

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08 Apr 2017, 09:47
1
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 ?

w* (w - 1) > 45
w >= 8?

(1) More than 1/2 of the 10 employees are women.
NS. w can be 6 through 10.

(2) The probability that both representatives selected will be men is less than 1/10.
m(m-1) < 9
m <=3, w can be 7 through 9. NS.

Combined,
m w
3 7 - does not satisfy.
2 8 - satisfies

NS.

E.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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28 Aug 2017, 18:22
Bunuel 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

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) More than 1/2 of the 10 employees are women --> $$w>5$$ not sufficient.

(2) The probability that both representatives selected will be men is less than 1/10 --> $$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> $$(10-w)(9-w)<9$$ --> $$w>6$$, not sufficient

(1)+(2) $$w>5$$ and $$w>6$$: $$w$$ can be 7, answer NO or more than 7, answer YES. Not sufficient.

You can use Combinations, to solve 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) More than 1/2 of the 10 employees are women --> $$w>5$$, not sufficient.

(2) The probability that both representatives selected will be men is less than 1/10 --> $$C^2_{(10-w)}$$ # 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

(1)+(2) $$w>5$$ and $$w>6$$: $$w$$ can be 7, answer NO or more than 7, answer YES. Not sufficient.

Hope it's clear.

Bunuel,
how do you go from $$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> $$(10-w)(9-w)<9$$
thank you. it would be nice if someone can explain this
Math Expert
Joined: 02 Sep 2009
Posts: 50610
Re: If 2 different representatives are to be selected at random  [#permalink]

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28 Aug 2017, 20:36
pclawong wrote:
Bunuel 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

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) More than 1/2 of the 10 employees are women --> $$w>5$$ not sufficient.

(2) The probability that both representatives selected will be men is less than 1/10 --> $$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> $$(10-w)(9-w)<9$$ --> $$w>6$$, not sufficient

(1)+(2) $$w>5$$ and $$w>6$$: $$w$$ can be 7, answer NO or more than 7, answer YES. Not sufficient.

You can use Combinations, to solve 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) More than 1/2 of the 10 employees are women --> $$w>5$$, not sufficient.

(2) The probability that both representatives selected will be men is less than 1/10 --> $$C^2_{(10-w)}$$ # 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

(1)+(2) $$w>5$$ and $$w>6$$: $$w$$ can be 7, answer NO or more than 7, answer YES. Not sufficient.

Hope it's clear.

Bunuel,
how do you go from $$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> $$(10-w)(9-w)<9$$
thank you. it would be nice if someone can explain this

It's done by multiplying both sides by 90.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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28 Aug 2017, 22:56
BANON 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.

An easy way to solve statement 2 might be:

Probability(both men)<1/10. Probability(neither men)= Probability(both women)< 1-1/10; Probability(both women) <9/10. (is it more than half? can be or cannot be)... Insufficient.
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Joined: 02 Sep 2009
Posts: 50610
Re: If 2 different representatives are to be selected at random  [#permalink]

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13 Dec 2017, 22:09
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-2-differe ... 68280.html
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Re: If 2 different representatives are to be selected at random &nbs [#permalink] 13 Dec 2017, 22:09

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