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# If 2 different representatives are to be selected at random

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

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27 Feb 2012, 09:11
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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 ?

(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.
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If 2 different representatives are to be selected at random [#permalink]

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27 Feb 2012, 09:15
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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

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.
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06 Aug 2012, 10:32
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Guys,

The way i attacked this problem was that quesn asks me if p(W,W) >1/2 ? Therefore,

(1) gives me women as 6, 7, 8, 9 (can't be 10).
Now, for 6 women, the probab would be p(W,W) = 6/10 * 5/9 = 1/3 .....less than half
Now, for 7 women, the probab would be p(W,W) = 7/10 * 6/9 = 7/15 .....less than half
Now, for 8 women, the probab would be p(W,W) = 8/10 * 7/9 = 28/45 .....more than half
Now, for 9 women, the probab would be p(W,W) = 9/10 * 8/9 = 4/5...more than half
Clearly, (1) is insufficient to answer... [eliminating A & D]

(2) gives me p(M,M) <1/10. Now, this is insuff. as it tells nothing abt p(W,W) unless i verify the above finding of (1) and club both
[B also eliminated, now contention is between C & E]
For 4 men, p(M,M) = 4/10 * 3/9 = 2/15 (grtr than 1/10) .....not valid
For 3 men, p(M,M) = 3/10 * 2/9 = 1/15 (less than 1/10) ...valid
For 2 men, p(M,M) = 2/10 * 1/9 = 1/45 (less than 1/10) ...valid

Thus, for 7W3M => p(W,W)<1/2 & p(M,M)<1/15
And, for 8W2M => p(W,W)>1/2 & p(M,M)< 1/45

So, combining 2 stmts is still insufficient to answer the original quesn. Hence, E has to be correct answer.
[PS: Initially i chose C, as i couldn't understand Bunuel's explanation above {which is a rarity }, but as i was posting this query, i realized that while choosing C, i didn't considered the 2 men case & that's why i chose wrongly ]

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

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08 May 2012, 01:41
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I made a silly mistake ...which i thought is worth sharing .
I thought that since number of women is greater than 5 so probability will be greater than ½ so A suff.
But A is insuff . coz the above statement will hold true only for a single event , but here 2 things are to be selected .
So better make equations and then derive condition.

Correct me if i am wrong.
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Re: Probability of desired outcome [#permalink]

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04 Nov 2012, 12:31
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JayGriffith8 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.

Please let me know if my methodology is correct. Say there are 6 women wouldn't the probability be 6/10*5/9? Yielding 1/3? I understand the maths behind this but I need to know a simple fast way of deriving an answer in this case. I feel like I'd be going back and forth with scenarios and eating too much time.

No. of Women X
Probability of selecting 2 Women >1 /2
X*(X-1)/2 / (10*9/2) > 1/2
X(X-1)>45
so X should be greater than 7.
or No. of Men should be less than 3

Statement A: Women can be 6,7,8,9,10 NS
Statement B: M(M-1)/10*9 < 1/10
M(M-1) < 9
M can be 0,1,2,3
NS.

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

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08 Apr 2017, 10:47
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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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11 Aug 2012, 05:00
how to reach the final statement

w> 6 from 2 we had (10-w)(9-w) < 9 , , why is w>6
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Re: If 2 different representatives are to be selected at random [#permalink]

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11 Aug 2012, 05:11
is plugging numbers only way to solve this quadratic inequality or do we have an algebric approach ?
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Re: If 2 different representatives are to be selected at random [#permalink]

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17 Aug 2012, 05:10
PUNEETSCHDV wrote:
how to reach the final statement

w> 6 from 2 we had (10-w)(9-w) < 9 , , why is w>6

If w=6 then (10-w)(9-w)=4*3=12>9 and if w=7, then (10-w)(9-w)=3*2=6<9. When we increase w, from 7 to 10, (10-w)(9-w) decreases so w can be 7, 8, 9 or 10.

PUNEETSCHDV wrote:
is plugging numbers only way to solve this quadratic inequality or do we have an algebric approach ?

We could expand (10-w)(9-w) and then solve quadratic inequality, but number plugging for this particular case is better.

Check this: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476 (solving quadratic inequalities)
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06 Nov 2012, 12:25
Bunuel wrote:
breakit 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.

Hi I couldn't able to understand the write uo which i have marked in red . Please help here..(not sure how to mark those picture stuff)

Check here: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622

Hope it helps.

Sorry , i am not understand how that formula stuff has been written(kind of lagging in that field)

[color=#ff0000] $$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> $$(10-w)(9-w)<9$$ --> $$w>6$$,

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07 Nov 2012, 05:44
breakit wrote:
Bunuel wrote:
breakit wrote:

Hi I couldn't able to understand the write uo which i have marked in red . Please help here..(not sure how to mark those picture stuff)

Check here: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622

Hope it helps.

Sorry , i am not understand how that formula stuff has been written(kind of lagging in that field)

[color=#ff0000] $$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> $$(10-w)(9-w)<9$$ --> $$w>6$$,

$$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> cross-multiply (multiply both parts by 10*9): $$(10-w)(10-w-1)<9$$ --> $$(10-w)(9-w)<9$$.

Hope it's clear.
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Re: If 2 different representatives are to be selected at random [#permalink]

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13 Mar 2013, 16:44
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.

This is extreme value problem

for p > 1/2 , p1 * p2 > 1/2 ie p1 or p2 > 1/4

We move with this further .
1. if P-new> 1/2 are women , then it can be say 1/2 (plus some point say .51) * .51 which is not sufficient

2. if P-men < 1/10 then P-women will be 9/10 ie there are many values between 1/4 and 9/10 which is not sufficient

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

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11 Dec 2016, 20:40
Bunuel
cant we do like this

P(both men)<0.1
1-P(both men)<0.9
P(both women)<0.9...so it can be >0.5 or <0.5

by combination also the same problem persist,hence E

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

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28 Aug 2017, 19:22
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

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

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28 Aug 2017, 21:36
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, 23:56
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] 28 Aug 2017, 23:56
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