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

Answer is E
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Re: If 2 different representatives are to be selected at random from a gro [#permalink]
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Re: If 2 different representatives are to be selected at random from a gro [#permalink]
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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 from a gro [#permalink]
Hey everyone - I have a question about this problem.

I understand how to solve it - that's not the issue at all (thank you so much Bunuel)
My issue is this - can you really do all the calculations that Bunuel showed in under 2 min?

I mean after I read all the explanations and timed myself on the problem I was on 2.07, given that I pretty much knew the answer, the calculations and everything that is given. In a realistic world, even knowing HOW to solve it, would probably take me >3.5min.

Questions:
1. Is that normal or am I just bad with arithmetic?
2. Is there a "shortcut" for this problem?
3. You probably noted the topic is called "combinatorics vs probability" - I am curious - who prefers which solution to this problem? I find the combinatorics one to be even longer, is that lack of practice? Does anyone use combi for this?
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Re: If 2 different representatives are to be selected at random from a gro [#permalink]
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imoi wrote:
Hey everyone - I have a question about this problem.

I understand how to solve it - that's not the issue at all (thank you so much Bunuel)
My issue is this - can you really do all the calculations that Bunuel showed in under 2 min?

I mean after I read all the explanations and timed myself on the problem I was on 2.07, given that I pretty much knew the answer, the calculations and everything that is given. In a realistic world, even knowing HOW to solve it, would probably take me >3.5min.

Questions:
1. Is that normal or am I just bad with arithmetic?
2. Is there a "shortcut" for this problem?
3. You probably noted the topic is called "combinatorics vs probability" - I am curious - who prefers which solution to this problem? I find the combinatorics one to be even longer, is that lack of practice? Does anyone use combi for this?


Usually, the combinatorics approach will be easier and more clear, especially in "order matter or not" cases. It might take a few seconds extra but is well worth it in my opinion.

If done properly (with variables and all), the question will take a long time - no getting away from it. Everyone (except Bunuel, perhaps!) will take quite a bit of time here. So in such a question one could work with numbers provided one can easily handle fractions.

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.

If no. of women is 6,
p = 6C2/10C2 = 6*5/9*10 = (6/9)*(1/2) which is less than 1/2 because when you multiply a number by something between 0 and 1, the number becomes smaller. SO 1/2 multiplied by 6/9 will give less than 1/2.
If number of women is 7,
p = 7*6/10*9 = 7/15 which is less than 1/2
If no of women is 9,
p = 9*8/10*9 = 8/10 (more than 1/2)
Not sufficient

(2) The probability that both representatives selected will be men is less than 1/10.
Say no of men is 3 (and number of women 7)
P of selecting both men = 3*2/10*9 = 1/15. This is less than 1/10. We have seen that with 7 women, p is less than 1/2.

Number of men can be 2 or 1 also since in those cases, the probability of selecting 2 men will be even less than 1/15. We have seen that when number of women is 9, p is greater than 1/2
Not sufficient.

Using both together, we have already seen 2 cases:
number of women = 7 and men = 3. p is less than 1/2 in this case
number of women = 9 and men = 1. p is more than 1/2 in this case

Hence answer will be (E)
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Re: If 2 different representatives are to be selected at random from a gro [#permalink]
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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.

Final Answer:

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

Answer E.

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.

Answer E.


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 from a gro [#permalink]
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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.

Answer E.

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.

Answer E.


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 from a gro [#permalink]
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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 from a gro [#permalink]
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Bunuel wrote:
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



Target question: Is the probability that both representatives selected will be women > 1/2?

This is a good candidate for rephrasing the target question.

Let's examine some scenarios and see which ones yield situation where the probability that both representatives selected will be women > 1/2
Scenario #1 - 5 women & 5 men: P(both selected people are women) = (5/10)(4/9) = 20/90 (NOT greater than 1/2)
Scenario #2 - 6 women & 4 men: P(both selected people are women) = (6/10)(5/9) = 30/90 (NOT greater than 1/2)
Scenario #3 - 7 women & 3 men: P(both selected people are women) = (7/10)(6/9) = 42/90 (NOT greater than 1/2)
Scenario #4 - 8 women & 2 men: P(both selected people are women) = (8/10)(7/9) = 56/90 (PERFECT - greater than 1/2)

IMPORTANT: So, if there are 8 or more women, the probability will be greater than 1/2
We can even REPHRASE the target question...
REPHRASED target question: Are there 8 or more women?

Statement 1: More than 1/2 of the 10 employees are women.
This is not enough information. Consider these two conflicting cases:
Case a: there are 7 women, in which case there are NOT 8 or more women
Case b: there are 8 women, in which case there ARE 8 or more women
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statement 2: The probability that both representatives selected will be men is less than 1/10.
This is not enough information. Consider these two conflicting cases:
Case a: there are 8 women & 2 men. Here P(both are men) = (2/10)(1/9) = 2/90, which is less than 1/2. In this case there ARE 8 or more women
Case b: there are 7 women & 3 men. Here P(both are men) = (3/10)(2/9) = 6/90, which is less than 1/2. In this case there are NOT 8 or more women
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Even when we combine the statements, we can see that it's possible to have 7 women in the group OR 8 women in the group.
Since we still cannot answer the REPHRASED target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

RELATED VIDEO ON REPHRASING THE TARGET QUESTION
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Re: If 2 different representatives are to be selected at random from a gro [#permalink]
GMATinsight : when we combine both statements selecting 7 women and 3 men proves to be true from both statements in the sense that probability of selecting 7 women is greater then half and probability of selecting 3 men is less than 1/10. can the answer be c??
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Re: If 2 different representatives are to be selected at random from a gro [#permalink]
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chanchal1311 wrote:
GMATinsight : when we combine both statements selecting 7 women and 3 men proves to be true from both statements in the sense that probability of selecting 7 women is greater then half and probability of selecting 3 men is less than 1/10. can the answer be c??


Hi chanchal1311,

When dealing with DS questions, you have to consider all of the 'restrictions' that the prompt places on you AND you have to make sure to answer the question that is ASKED. By extension, you have to be thorough with your work - and define whether there's more than one possible situation that fits everything you know or not.

Each Fact is insufficient on its own. When combining the Facts, we have examples that fit both Facts:

If the number of women is 7, then the answer to the question is NO.
If the number of women is 10, then the answer to the question is YES.
Thus, even when we combine the Facts, we still end up with INSUFFICIENT as the answer.

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Re: If 2 different representatives are to be selected at random from a gro [#permalink]
Hi guys Bunuel VeritasKarishma, it would be very helpful if anyone could please let me know where I went wrong.

P(zero women)+ P(one woman) + P(two women) = 1
As per statement 2, P(zero women) < 1/10.
P(one woman) = (w/10) * ((10-w)/9) where w is the number of women in the group, so this equals (10w-w^2)/90.
P(two women) = (w/10)*((w-1)/9) = (w^2-w)/90
Now, (10w-w^2)/90 + (w^2-w)/90 + (something less than 1/10) = 1
On simplification, w/10 = 1- (something less than 1/10) => w/10 = something greater than (9/10) => w is something greater than 9 therefore, p>1/2.
B seems to be the answer.
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Re: If 2 different representatives are to be selected at random from a gro [#permalink]
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RishavRaj wrote:
Hi guys Bunuel VeritasKarishma, it would be very helpful if anyone could please let me know where I went wrong.

P(zero women)+ P(one woman) + P(two women) = 1
As per statement 2, P(zero women) < 1/10.
P(one woman) = (w/10) * ((10-w)/9) where w is the number of women in the group, so this equals (10w-w^2)/90.
P(two women) = (w/10)*((w-1)/9) = (w^2-w)/90
Now, (10w-w^2)/90 + (w^2-w)/90 + (something less than 1/10) = 1
On simplification, w/10 = 1- (something less than 1/10) => w/10 = something greater than (9/10) => w is something greater than 9 therefore, p>1/2.
B seems to be the answer.



The highlighted part is incorrect.
P(one women) = P(Man - Woman) + P(Woman - Man)

(w/10) * ((10-w)/9) = P(Woman - Man)
This needs to be multiplied by 2 to account for the probability of getting Man- Woman.

Basically,

MM + MW + WM + WW = 1 (using probability method)
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