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If 2 different representatives are to be selected at random

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

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

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New post 08 Apr 2017, 09:47
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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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New post 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.

Answer E.

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.

Answer E.

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

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

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New post 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.

Answer E.

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.

Answer E.

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

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Re: If 2 different representatives are to be selected at random   [#permalink] 13 Dec 2017, 22:09

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