Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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
Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.
We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.
If 2 different representatives are to be selected at random [#permalink]
Show Tags
04 Mar 2014, 00:17
15
13
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.
Re: If 2 different representatives are to be selected at random [#permalink]
Show Tags
06 Mar 2014, 11:42
3
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.
Re: If 2 different representatives are to be selected at random [#permalink]
Show Tags
07 Mar 2014, 01:59
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
St1. Clearly answers so sufficient 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
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
Re: If 2 different representatives are to be selected at random [#permalink]
Show Tags
07 Mar 2014, 08:13
2
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
St1. Clearly answers so sufficient 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).
Re: If 2 different representatives are to be selected at random [#permalink]
Show Tags
08 Mar 2014, 05:14
1
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
Status: suffer now and live forever as a champion!!!
Joined: 01 Sep 2013
Posts: 135
Location: India
Dheeraj: Madaraboina
GPA: 3.5
WE: Information Technology (Computer Software)
Re: If 2 different representatives are to be selected at random [#permalink]
Show Tags
03 Jun 2014, 06:53
1
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
Re: If 2 different representatives are to be selected at random [#permalink]
Show Tags
08 Jun 2015, 18:41
2
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.
Re: If 2 different representatives are to be selected at random [#permalink]
Show Tags
24 Jun 2015, 03:27
Hi,
I have a question concerning statement 2. I assumed that "probability that both are men is less than 10%" can be equal to (1-p) of the original statement "p(both women) > 50%?"
Re: If 2 different representatives are to be selected at random [#permalink]
Show Tags
24 Jun 2015, 11:04
1
Hi noTh1ng,
Here's a hint (so that you can keep working on this): there are other possibilities besides "both men" and "both women"; what are they and how would those OTHER possibilities impact your calculation?
Re: If 2 different representatives are to be selected at random [#permalink]
Show Tags
28 Jul 2016, 06:43
1
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?
_________________
You have to dig deep and find out what it takes to reshuffle the cards life dealt you
Re: If 2 different representatives are to be selected at random [#permalink]
Show Tags
28 Jul 2016, 06:52
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.
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.
_________________
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
_________________
Brent Hanneson – Founder of gmatprepnow.com
gmatclubot
Re: If 2 different representatives are to be selected at random
[#permalink]
24 Apr 2018, 11:09