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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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03 Mar 2014, 23:17
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

Data Sufficiency
Question: 122
Category: Arithmetic Probability
Page: 162
Difficulty: 650

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

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03 Mar 2014, 23:17
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SOLUTION

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 ?

The probability of selecting 2 women out of 10 people is $$\frac{w}{10}*\frac{w-1}{9}$$.

The question asks whether $$\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}$$ --> is $$w(w-1)>45$$? --> 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:

$$C^2_w$$ the number of selections of 2 women out of $$w$$ employees;

$$C^2_{10}$$ the total number of selections of 2 representatives out of 10 employees.

The question asks whether $$\frac{C^2_w}{C^2_{10}}>\frac{1}{2}$$ --> is $$\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}$$? --> is $$w(w-1)>45$$? --> 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 --> $$C^2_{(10-w)}$$ # 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.

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

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06 Mar 2014, 10:42
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I get E. Please let me know if the work is correct.

(1) There are 6-10 women.

There are 45 ways to choose the team. 10C2 = 45. If there are 6 women, Pr(W, W) = 6C2/10C2 = 15/45 = 1/3 < 1/2 No
If there are 10 women, Pr (W, W) = 1 Yes

Therefore (1) is not sufficient. Eliminate A and D.

(2) Pr (M, M) < 1/10

xC2/45 < 1/10

xC2 < 4.5

There can be at most 3 men. 4C2 = 6 > 4.5, but 3C2 = 3 < 4.5

If there are at most 3 men, there are at least 7 women.

7C2/45 = 21/45 < 1/2 No
10C2/45 = 1 > 1/2 Yes

Eliminate B

(1/2) Statement 2 is just a subset of the possibilities for Statement 1. Still not sufficient. The answer is E.

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

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07 Mar 2014, 00:59
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

Sol: Probability p can be defined as no of favourable cases (selecting 2 women)/ Total no. of possible cases

we need to know is p>1/2 Basically we need to check whether the number of women in the group is more than 5

St 2 The probability of both rep selected to be men is less than 1/10
Assume there are 2 men in the group and 8 women. Now probability of selecting 2 men will be 2C2/ 10C2 or 1/45 <1/10 (or 0.1)
for 3 men and 7 women , Probability will be 3C2/10C2 or 3/45 or 1/15 ie. 1/15 <0.1
for 4 men and 6 women, Prob. will be 4C2/10C2 oe 2/15 >0.1...Thus we can conclude that no. of women are more than 5

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

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07 Mar 2014, 07:13
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WoundedTiger 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

Sol: Probability p can be defined as no of favourable cases (selecting 2 women)/ Total no. of possible cases

we need to know is p>1/2 Basically we need to check whether the number of women in the group is more than 5

St 2 The probability of both rep selected to be men is less than 1/10
Assume there are 2 men in the group and 8 women. Now probability of selecting 2 men will be 2C2/ 10C2 or 1/45 <1/10 (or 0.1)
for 3 men and 7 women , Probability will be 3C2/10C2 or 3/45 or 1/15 ie. 1/15 <0.1
for 4 men and 6 women, Prob. will be 4C2/10C2 oe 2/15 >0.1...Thus we can conclude that no. of women are more than 5

Hence Ans is D

I believe we need to check statement 1. For example, if there are 6 women for 2 spaces, we get Pr (W, W) = 1/3:

The probability of picking the second woman is not independent of the probability of picking the first.
There are two open spots:

______ ______

6/10 * 5/9

30/90

1/3 < 1/2

The probability the first spot goes to a woman is the proportion of women in the group. The probability the second spot goes to a woman is the proportion of women from the remaining group. We are selecting "without replacement."

At the other extreme, if the group is composed entirely of women (I think one "trick" is that we are never told this group is co-ed), it is a certainty that the committee will be all-female. Pr (W, W) = 1

Thus, we cannot determine whether p > 1/2 from Statement 1.

The reasoning for Statement 2 is correct in that there are at most 3 men. The smallest number of women must be 7.

Check this case:

______ ______

7/10 * 6/9

14/30 < 1/2

Recognize again that there can be zero men in the group. In that case Pr (W, W) = 1 > 1/2

Thus we cannot determine sufficiency based on Statement 2.

Combining the statements yields no new info because (2) is just a more restricted version of (1).

The answer should therefore be E.

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

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08 Mar 2014, 04:14
1
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Statemenet 1: if P=6/10*5/9=30/90, i.e.<1/2, but if 10/10*9/9=1, i.e.>1/2, NOT SUFFICIENT
Statement 2: x/10*y/9<1/10 => xy<9 with multiplying consecutives, so it can be 3 men and 7 women =>7/10*6/9=42/90, i.e.<1/2 but it can be 2 men and 8 women =>8/10*7/9=56/90, i.e>1/2, NOT SUFFICIENT
Statements 1 and 2: It just gives us that women number cannot be less than 7, but again it is NOT SUFFICIENT

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

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08 Mar 2014, 13:00
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Re: If 2 different representatives are to be selected at random [#permalink]

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03 Jun 2014, 05:53
stmt1: More than 1/2 of 10 employees are women.

case 1: 6 women , 4 men
probability of selecting 2 women is (6/10)*(5/9) = 1/3<1/2 (true)
case 2: 7 women , 3 men
probability of selecting 2 women is (7/10)*(6/9) = 7/15<1/2 (true)
case 3: 8 women , 2 men
probability of selecting 2 women is (8/10) * (7/9) = 28/45>1/2 (false)

Hence Insufficient

stmt2: :The probability that both representatives selected will be men is less than 1/10

case1 : 8 women, 2 men
probability of selecting 2 men is (2/10)*(1/9) = 1/45 <1/10 (true)
probability of selecting 2 women is (8/10) * (7/9) = 28/45>1/2 (false)
case 2 : 7 women , 3 women
probability of selecting 2 men is (3/10)*(2/9) = 1/15 <1/10 (true)
probability of selecting 2 women is (7/10)*(6/9) = 7/15<1/2 (true)
Hence Insufficient.

Both statements combined is not sufficient.
Hence E .
Hope this helps

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

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08 Jun 2015, 17:41
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Expert's post
Hi All,

We're told that there are 10 employees (men and women). We're asked if the probability of selected 2 women is > 1/2.

Fact 1: More than half are women.

This means the number of women COULD BE 6, 7, 8, 9 or 10.

IF...there are 7 women...
the probability of selecting 2 women is (7/10)(6/9) = 42/90 = LESS than 1/2 and the answer to the question is NO.

IF....there are 10 women...
the probability of selecting 2 women is 100% and the answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: The probability of getting 2 men is < 1/10.

IF...there are 3 men, then the probability of selecting 2 men is (3/10)(2/9) = 6/90 = 1/15 which is less than 1/10
In this situation, there would be 7 women....and the answer to the question is NO.

IF....there are 0 men, then the probability of selecting 2 men is 0%.
In this situation, there would be 10 women....and the answer to the question is YES.
Fact 2 is INSUFFICIENT.

Combining Facts, we have TESTs that fit "both" Facts:
If the number of women is 7, then the answer is NO.
If the number of women is 10, then the answer is YES.
Combined, INSUFFICIENT.

[Reveal] Spoiler:
E

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

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28 Jul 2016, 05:43
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Bunuel wrote:
SOLUTION

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 ?

The probability of selecting 2 women out of 10 people is $$\frac{w}{10}*\frac{w-1}{9}$$.

The question asks whether $$\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}$$ --> is $$w(w-1)>45$$? --> 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:

$$C^2_w$$ the number of selections of 2 women out of $$w$$ employees;

$$C^2_{10}$$ the total number of selections of 2 representatives out of 10 employees.

The question asks whether $$\frac{C^2_w}{C^2_{10}}>\frac{1}{2}$$ --> is $$\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}$$? --> is $$w(w-1)>45$$? --> 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 --> $$C^2_{(10-w)}$$ # 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.

Hi :

How did you solve for (w -10)*(w - 9) < 0 ?
w^2 - 19w +81 < 0 ------> what are the factors of this?
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Re: If 2 different representatives are to be selected at random [#permalink]

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28 Jul 2016, 05:52
aniketm.87@gmail.com wrote:
Bunuel wrote:
SOLUTION

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 ?

The probability of selecting 2 women out of 10 people is $$\frac{w}{10}*\frac{w-1}{9}$$.

The question asks whether $$\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}$$ --> is $$w(w-1)>45$$? --> 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:

$$C^2_w$$ the number of selections of 2 women out of $$w$$ employees;

$$C^2_{10}$$ the total number of selections of 2 representatives out of 10 employees.

The question asks whether $$\frac{C^2_w}{C^2_{10}}>\frac{1}{2}$$ --> is $$\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}$$? --> is $$w(w-1)>45$$? --> 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 --> $$C^2_{(10-w)}$$ # 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.

Hi :

How did you solve for (w -10)*(w - 9) < 0 ?
w^2 - 19w +81 < 0
------> what are the factors of this?

We could expand (10-w)(9-w) and then solve quadratic inequality, but number plugging for this particular case is better.
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Re: If 2 different representatives are to be selected at random [#permalink]

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