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A certain business school has 500 students, and the law [#permalink]
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10 Nov 2012, 04:30
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A certain business school has 500 students, and the law school at the same university has 800 students. Among these students, there are 30 sibling pairs consisting of 1 business student and 1 law student. If 1 student is selected at random from both schools, what is the probability that a sibling pair is selected? A. \(\frac{3}{40000}\) B. \(\frac{3}{20000}\) C. \(\frac{3}{4000}\) D. \(\frac{9}{400}\) E. \(\frac{6}{130}\) However, I selected B. Since order doesn't matter, I added \(\frac{3}{40000}\) to \(\frac{3}{40000}\) and got \(\frac{3}{20000}\). Could someone explain why I am wrong??
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Last edited by Bunuel on 10 Nov 2012, 04:38, edited 1 time in total.
Renamed the topic and edited the question.



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Re: A certain business school has 500 students, and the law [#permalink]
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10 Nov 2012, 04:42



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Re: A certain business school has 500 students, and the law [#permalink]
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10 Nov 2012, 04:58
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My weakest link in GMAT prepping. probability of selecting 1 student from Harvards Business School1/500 probability of selecting 1 student from Harvards Law School1/800 probability that these two students are siblings(1/500 * 1/800) since there are 30 siblings, hence (1/500 * 1/800)*30. Answer 3/40000 Hope that helps
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Re: A certain business school has 500 students, and the law [#permalink]
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In my opinion answer is A, probability of chosing one subling from business school is 30/500 we multiply it to 1/800 the probability of choosing a subling from law school, for the one that we have already chosen from business school, since there is only one in law school so the probability is 1/800. Overall : 30/500*1/800=3/40000 (A). But i think this question need to be modified slightly EACH should be added to the phrase "there are 30 sibling pairs, each consisting of 1 business student and 1 law student" Otherwise there could be different understanding. Hope that helps
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Re: A certain business school has 500 students, and the law [#permalink]
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12 Nov 2012, 04:34
No. of all possible pairs: 1/(500x800) = 1/400,000 No. of all desired sibling pairs: 30 30/400,000 = 3/40,000 Answer: A
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Re: A certain business school has 500 students, and the law [#permalink]
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05 Dec 2012, 16:09
MacFauz wrote: A certain business school has 500 students, and the law school at the same university has 800 students. Among these students, there are 30 sibling pairs consisting of 1 business student and 1 law student. If 1 student is selected at random from both schools, what is the probability that a sibling pair is selected? A. \(\frac{3}{40000}\) B. \(\frac{3}{20000}\) C. \(\frac{3}{4000}\) D. \(\frac{9}{400}\) E. \(\frac{6}{130}\) However, I selected B. Since order doesn't matter, I added \(\frac{3}{40000}\) to \(\frac{3}{40000}\) and got \(\frac{3}{20000}\). Could someone explain why I am wrong?? A certain business school has 500 students, and the law school at the same university has 800 students. Among these students, there are 30 sibling pairs consisting of 1 business student and 1 law student. If 1 student is selected at random from both schools, what is the probability that a sibling pair is selected? A) 3/40,000 B) 3/20,000 C) 3/4,000 D) 9/400 E) 6/130 I have a question since I choose answer choice B My reasoning was that we can choose one business school or law school sibling, meaning we can choose business school student first or law school student first: 30/500*1/800= 3/40,000 or 30/800*1/500= 3/40,000 then add up totals 3/40,000+3/40,000=3/20,000. Because in some questions we do add totals but why don't we add up totals in this question? Thanks!!!



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Re: A certain business school has 500 students, and the law [#permalink]
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Let me put it this way. Ben and Lilian are siblings. Ben is studying Business and Lilian is studying Law.  Possibility of selecting Ben = 1/500  Possibility of selecting Lilian = 1/800 So the possibility of selecting this specific sibling pair is (1/500) * (1/800) = 1/400,000 Now imagine this concept with the total 30 sibling pairs. (1/500) * (1/800) * 30 = 30/400,000 = 3/40,000
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Re: A certain business school has 500 students, and the law [#permalink]
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19 Jul 2015, 17:37
Probability of selecting one sibling pair from each school = \(30(1/500 * 1/800) = 30(1/500*800) = 30/400000 = 3/40000\). Ans (A).
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Re: A certain business school has 500 students, and the law [#permalink]
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20 Jul 2015, 16:11
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Since we do not care which pair, the chance of selecting one of them from the business school is 30/500. The chance of selecting that specific student's sibling from the law school class is 1/800.
We multiply them together 30/500 * 1/800 = 30/400,000 = 3/40,000



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Re: A certain business school has 500 students, and the law [#permalink]
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22 Jul 2015, 02:13
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1st  you need to pick on from any school, who has a sibling P(l) = 15/500 P(b) = 15/800
2nd  you need to pick his pair sibling from the other school P(bl) = 1/800 P9lb) = 1/500
3rd, you need to multiply resepectively ("AND"  for example  both P(l) and P(bl) needed) and sum the result ("OR", one of the options needed)
1. 15/500*1/800 = 15/400,000 2. 15/800*1/500 = 15/400,000
1+2. = 15/400,000 +15/400,000 = 30/400,000 = 3/40,000



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Re: A certain business school has 500 students, and the law [#permalink]
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07 Dec 2016, 19:29
(30/500)(1/800) = 3/40,000
A.



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Re: A certain business school has 500 students, and the law [#permalink]
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10 Dec 2016, 23:54
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MacFauz wrote: A certain business school has 500 students, and the law school at the same university has 800 students. Among these students, there are 30 sibling pairs consisting of 1 business student and 1 law student. If 1 student is selected at random from both schools, what is the probability that a sibling pair is selected? A. \(\frac{3}{40000}\) B. \(\frac{3}{20000}\) C. \(\frac{3}{4000}\) D. \(\frac{9}{400}\) E. \(\frac{6}{130}\) However, I selected B. Since order doesn't matter, I added \(\frac{3}{40000}\) to \(\frac{3}{40000}\) and got \(\frac{3}{20000}\). Could someone explain why I am wrong?? Approach with application of combinatorics. Let’s start with business school. We can choose one sibling from \(30\) in \(_{30}C_1\) way. In general we can choose 1 person from 500 students in school in \(_{500}C_1\) way. After we have chosen one sibling from business school there is only one option to choose brother/sister from law school and we have: \(\frac{_{30}C_1}{_{500}C_1} * \frac{1}{_{800}C_1} = \frac{30}{500*800}\) But we chose to start from business school arbitrarily. We can also start choosing from law school first: \(\frac{_{30}C_1}{_{800}C_1} * \frac{1}{_{500}C_1} = \frac{30}{800*500}\) The probability of choosing one school from available two is \(\frac{1}{2}\) and we have: \(\frac{1}{2} * (\frac{30}{500*800} + \frac{30}{800*500}) = \frac{1}{2} * \frac{60}{500*800} = \frac{3}{4000}\) Answer A.



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Re: A certain business school has 500 students, and the law [#permalink]
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18 Oct 2017, 13:33
Hi, can someone please clarify why does order matter ? Why don’t we sum up the probabilities which would result in b as the answer that is, first we select a sibling from law school and then from business school and other situation is first from business school and then law school.
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Re: A certain business school has 500 students, and the law [#permalink]
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24 Oct 2017, 05:12
MacFauz wrote: A certain business school has 500 students, and the law school at the same university has 800 students. Among these students, there are 30 sibling pairs consisting of 1 business student and 1 law student. If 1 student is selected at random from both schools, what is the probability that a sibling pair is selected?
A. \(\frac{3}{40000}\)
B. \(\frac{3}{20000}\) C. \(\frac{3}{4000}\)
D. \(\frac{9}{400}\)
E. \(\frac{6}{130}\) The probability of selecting from the business school any one sibling from the 30 sibling pairs is 30/500. Once that person is selected, the probability of selecting his or her sibling from the law school is 1/800; thus, the probability of a selecting a sibling pair is: 30/500 x 1/800 = 3/50 x 1/800 = 3/40000 Alternatively, the probability of selecting from the law school any one sibling from the 30 sibling pairs is 30/800. Once that person is selected, the probability of selecting his or her sibling from the business school is 1/500; thus, the probability of a selecting a sibling pair is: 30/800 x 1/500 = 3/80 x 1/500 = 3/40000 Answer: A
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Re: A certain business school has 500 students, and the law [#permalink]
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26 Oct 2017, 11:34
ScottTargetTestPrep wrote: MacFauz wrote: A certain business school has 500 students, and the law school at the same university has 800 students. Among these students, there are 30 sibling pairs consisting of 1 business student and 1 law student. If 1 student is selected at random from both schools, what is the probability that a sibling pair is selected?
A. \(\frac{3}{40000}\)
B. \(\frac{3}{20000}\) C. \(\frac{3}{4000}\)
D. \(\frac{9}{400}\)
E. \(\frac{6}{130}\) The probability of selecting from the business school any one sibling from the 30 sibling pairs is 30/500. Once that person is selected, the probability of selecting his or her sibling from the law school is 1/800; thus, the probability of a selecting a sibling pair is: 30/500 x 1/800 = 3/50 x 1/800 = 3/40000 Alternatively, the probability of selecting from the law school any one sibling from the 30 sibling pairs is 30/800. Once that person is selected, the probability of selecting his or her sibling from the business school is 1/500; thus, the probability of a selecting a sibling pair is: 30/800 x 1/500 = 3/80 x 1/500 = 3/40000 Answer: A Hi ScottTargetTestPrepwe have 2 cases, first: we select first student from business school and second from law school and second vise versa why we don't add two posibilities?



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Re: A certain business school has 500 students, and the law [#permalink]
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05 Nov 2017, 06:29
soodia wrote: ScottTargetTestPrep wrote: MacFauz wrote: A certain business school has 500 students, and the law school at the same university has 800 students. Among these students, there are 30 sibling pairs consisting of 1 business student and 1 law student. If 1 student is selected at random from both schools, what is the probability that a sibling pair is selected?
A. \(\frac{3}{40000}\)
B. \(\frac{3}{20000}\) C. \(\frac{3}{4000}\)
D. \(\frac{9}{400}\)
E. \(\frac{6}{130}\) The probability of selecting from the business school any one sibling from the 30 sibling pairs is 30/500. Once that person is selected, the probability of selecting his or her sibling from the law school is 1/800; thus, the probability of a selecting a sibling pair is: 30/500 x 1/800 = 3/50 x 1/800 = 3/40000 Alternatively, the probability of selecting from the law school any one sibling from the 30 sibling pairs is 30/800. Once that person is selected, the probability of selecting his or her sibling from the business school is 1/500; thus, the probability of a selecting a sibling pair is: 30/800 x 1/500 = 3/80 x 1/500 = 3/40000 Answer: A Hi ScottTargetTestPrepwe have 2 cases, first: we select first student from business school and second from law school and second vise versa why we don't add two posibilities? Same doubt here...any of the expert might shed some light on why we don't add up? ScottTargetTestPrep Bunuel Thanks a lot for your support



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Re: A certain business school has 500 students, and the law [#permalink]
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07 Nov 2017, 21:18
probability of selecting sibling 30 students from b school 30/500 probability of selecting sibling 30 students from law school 30/800 probability that selected sibling students are members of the same sibling pair 1/30 (30/500)*(30/800)*(1/30) = 3/40000



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Re: A certain business school has 500 students, and the law [#permalink]
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08 Nov 2017, 04:54
delid wrote: Same doubt here...any of the expert might shed some light on why we don't add up? ScottTargetTestPrep Bunuel Thanks a lot for your support It is because we are picking two people from two different sets so order in which they are picked does not matter. Therefore, all you need to do is pick one from each set. Whether you pick the business fellow first or the lawyer first is immaterial. Hence you don't need to account for both cases. I have taken this concept in detail here: https://www.veritasprep.com/blog/2013/0 ... ermatter/It tells you the kind of questions in which order matters and the kind in which order does not matter.
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