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Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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29 Jul 2016, 03:39
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Company X employs more than 5 people, and will choose 2 employees at random to send to a business meeting. Is the probability that Company X will select 2 women greater than 0.5? (1) More than 70% of the employees of Company X are women. (2) Company X employs more than 12 people in total.
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Last edited by Bunuel on 29 Jul 2016, 03:55, edited 1 time in total.
Renamed the topic and edited the question.



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Re: Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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29 Jul 2016, 07:08
St1: More than 70% of the employees of Company X are women. Consider the number of employees = 10 W > 7 P(2 women) > (7C2)/(10C2) = 42/90 = 7/15 Probability can be greater or lesser than 0.5 Not Sufficient
St2: Number of employees > 12 > No information about the number of women. If the number of women employees = 1 then probability = 0 If all the employees are women then probability = 100% Not Sufficient
Combining St1 and St2: Number of employees > 12 and W > 0.7*(Number of employees) Let number of employees = 20 W > 14 P(2 women) > (14C2)/(20C2) = 14*13/20*19 = 91/190 Probability can be greater or lesser than 0.5 Not Sufficient
Answer: E



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Re: Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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29 Jul 2016, 08:25
I feel answer is A  Here is my argument
lets assume there are 100 employees
probability of selecting two woman = Probability of selecting 1st woman and Probability of selecting 2nd woman = 70/100*(69/99)
even if the number of employees is not know we can say
= [.7x/x] *[(.7x1)/x1]
which implies
.7* some fraction less than .7 <.5
please correct me if I am wrong hence it probability of of two women selection will never cross .5 if there are 70% woman in organization



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Re: Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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30 Jul 2016, 01:26
utkarsh240884 wrote: I feel answer is A  Here is my argument
lets assume there are 100 employees
probability of selecting two woman = Probability of selecting 1st woman and Probability of selecting 2nd woman = 70/100*(69/99)
even if the number of employees is not know we can say
= [.7x/x] *[(.7x1)/x1]
which implies
.7* some fraction less than .7 <.5
please correct me if I am wrong hence it probability of of two women selection will never cross .5 if there are 70% woman in organization You have written ".7* some fraction less than .7 <.5". I think you meant 0.5<0.7 See, we have some fraction less than 0.7 but that doesn't guarantee if it would be greater than 0.5. Thus INSUFFICIENT.
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Re: Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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30 Jul 2016, 08:25
Vyshak wrote: St1: More than 70% of the employees of Company X are women. Consider the number of employees = 10 W > 7 P(2 women) > (7C2)/(10C2) = 42/90 = 7/15 Probability can be greater or lesser than 0.5 Not Sufficient
St2: Number of employees > 12 > No information about the number of women. If the number of women employees = 1 then probability = 0 If all the employees are women then probability = 100% Not Sufficient
Combining St1 and St2: Number of employees > 12 and W > 0.7*(Number of employees) Let number of employees = 20 W > 14 P(2 women) > (14C2)/(20C2) = 14*13/20*19 = 91/190 Probability can be greater or lesser than 0.5 Not Sufficient
Answer: E Sir i BELIEVE ANS SHOULD BE A If you take any number greater than 5 and has women percantage 70% or more probability will always be less than .5



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Re: Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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30 Jul 2016, 08:35
Shivam2190 wrote: Vyshak wrote: St1: More than 70% of the employees of Company X are women. Consider the number of employees = 10 W > 7 P(2 women) > (7C2)/(10C2) = 42/90 = 7/15 Probability can be greater or lesser than 0.5 Not Sufficient
St2: Number of employees > 12 > No information about the number of women. If the number of women employees = 1 then probability = 0 If all the employees are women then probability = 100% Not Sufficient
Combining St1 and St2: Number of employees > 12 and W > 0.7*(Number of employees) Let number of employees = 20 W > 14 P(2 women) > (14C2)/(20C2) = 14*13/20*19 = 91/190 Probability can be greater or lesser than 0.5 Not Sufficient
Answer: E Sir i BELIEVE ANS SHOULD BE A If you take any number greater than 5 and has women percantage 70% or more probability will always be less than .5 Take number of employees = 1000; number of women = 701 Probability = 701C2/1000C2 = 701*700/(1000*999) = 490700/999000 = 4907/9990 < 0.5 Answer cannot be A.



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Re: Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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04 Aug 2016, 07:45
@ abhimahna wrote: utkarsh240884 wrote: I feel answer is A  Here is my argument
lets assume there are 100 employees
probability of selecting two woman = Probability of selecting 1st woman and Probability of selecting 2nd woman = 70/100*(69/99)
even if the number of employees is not know we can say
= [.7x/x] *[(.7x1)/x1]
which implies
.7* some fraction less than .7 <.5
please correct me if I am wrong hence it probability of of two women selection will never cross .5 if there are 70% woman in organization You have written ".7* some fraction less than .7 <.5". I think you meant 0.5<0.7 See, we have some fraction less than 0.7 but that doesn't guarantee if it would be greater than 0.5. Thus INSUFFICIENT. Thanks Abhimaan for input  but I mean .7 multiplied by some fraction less than .7 which will be <.5



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Re: Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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04 Aug 2016, 10:01
utkarsh240884 wrote: @ abhimahna wrote: utkarsh240884 wrote: I feel answer is A  Here is my argument
lets assume there are 100 employees
probability of selecting two woman = Probability of selecting 1st woman and Probability of selecting 2nd woman = 70/100*(69/99)
even if the number of employees is not know we can say
= [.7x/x] *[(.7x1)/x1]
which implies
.7* some fraction less than .7 <.5
please correct me if I am wrong hence it probability of of two women selection will never cross .5 if there are 70% woman in organization You have written ".7* some fraction less than .7 <.5". I think you meant 0.5<0.7 See, we have some fraction less than 0.7 but that doesn't guarantee if it would be greater than 0.5. Thus INSUFFICIENT. Thanks Abhimaan for input  but I mean .7 multiplied by some fraction less than .7 which will be <.5 Bro, How can be you be assured that that 0.7 * some fraction less than 0.7 would in turn be less than 0.5?
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Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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22 Sep 2016, 06:36
Keats wrote: Company X employs more than 5 people, and will choose 2 employees at random to send to a business meeting. Is the probability that Company X will select 2 women greater than 0.5?
(1) More than 70% of the employees of Company X are women. (2) Company X employs more than 12 people in total. 1. More than 70% of the employees of Company X are women.If there are 6 people and all of them are women, than answer is YES. But if there are very large number of employess and the quantity of women is 70%+1 than the chance will be closer to 0.7*0.7=.49 and the answer is NO (the more quantity of workers the more chance is close to 0.49) NOT SUFFICIENT2.Company X employs more than 12 people in total.All of them can be women (than YES), all of them can be men (than NO) NOT SUFFICIENT1+2.if there are very large number of employess and the quantity of women is 70%+1 than the chance will be closer to 0.7*0.7=.49 and the answer is NOIf there are 13 people and all of them are women, than answer is YES. Therefore, NOT SUFFICIENT, E
Last edited by dimka on 22 Sep 2016, 22:16, edited 1 time in total.



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Re: Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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22 Sep 2016, 11:37
I'm confused as to why we're using combinatorics vs. probability.
I saw in the first response it assumed we have 10 people, but it says that there are MORE than 70% women so minimum 8.
Wouldn't we say for the first option 8/10 then the second 7/9 and multiply?
I see why it is E and the example of 1000 employees make sense...how are we to know this without trying multiple options?



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Re: Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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22 Sep 2016, 13:04
joannaecohen wrote: I'm confused as to why we're using combinatorics vs. probability.
I saw in the first response it assumed we have 10 people, but it says that there are MORE than 70% women so minimum 8.
Wouldn't we say for the first option 8/10 then the second 7/9 and multiply?
I see why it is E and the example of 1000 employees make sense...how are we to know this without trying multiple options? To find the answer we can use differnt methods. such as combinatorics and probability, but I am a little disagree with the first response. If we have 10 people, and more than 70 % of them are women, then the minimum number of women is 8. In this case chance that we select a woman during the first try is 8/10, and chance that we after that select a woman during the second try is 7/9. So the total probability is 8/10*7/9=28/45 or about 62 % We can find this answer using combinatorics (8C2)/(10C2)=28/45 or about 62 %. On my opinion, the main problem is to pick up the number which gives chance less than 50%. You can use example with 1000 workers, but in this case you have to prove that (701/1000)*(700/999)>1/2. I can't do so during 2 minutes that's why I sugest using very very large number. Large enough that you could just multiply 0.7*0.7 and get 0.49, showing that there can be the case when probability is less than 50 %. Hopes it's clear. D.



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Re: Company X employs more than 5 people, and will choose 2 employees at [#permalink]
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23 Feb 2017, 05:12
Prompt analysis Company X has more than 5 employs and 2 employs were selected.
Super set The answer will be in the range 0 to 1.
Translation In order to find the answer, we need: 1# exact number of men and women 2# rough idea about the range and percentage of men and women.
Statement analysis
St1: More than 70% of the employees of Company X are women. Consider the number of employees = 10 W > 7 P(2 women) > (7C2)/(10C2) = 42/90 = 7/15 Probability can be greater or lesser than 0.5 Not Sufficient
St2: Number of employees > 12 > No information about the number of women. If the number of women employees = 1 then probability = 0 If all the employees are women then probability = 100% Not Sufficient
Combining St1 and St2: Number of employees > 12 and W > 0.7*(Number of employees) Let number of employees = 20 W > 14 P(2 women) > (14C2)/(20C2) = 14*13/20*19 = 91/190 Probability can be greater or lesser than 0.5 Not Sufficient
Option E




Re: Company X employs more than 5 people, and will choose 2 employees at
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