Hello,
For the question below, why isn't the answer: (30/500 * 1/800) + (30/800 * 1/500). We can the first student either from business school or law school...
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 each school, what is the probability that a sibling pair is selected?
A) 3/40000
B) 3/20000
C) 3/4000
D) 9/400
E) 6/130
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Combinations problem
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topspin360
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GMATGuruNY
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You're double-counting the sibling pairs.topspin360 wrote:Hello,
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 each school, what is the probability that a sibling pair is selected?
A) 3/40000
B) 3/20000
C) 3/4000
D) 9/400
E) 6/130
OAA
For the question above, why isn't the answer: (30/500 * 1/800) + (30/800 * 1/500). We can the first student either from business school or law school...
Since each pair consists of 1 junior and 1 senior, it doesn't matter whether we select first from the junior class and then from the senior class or first from the senior class and then from the junior class: either way, we'll be counting the SAME sibling pairs.
An alternate approach:
P(sibling pair) = (total number of sibling pairs)/(total number of possible pairs).
Total number of possible pairs:
There are 500 juniors and 800 seniors.
Total number of ways to combine 1 junior with 1 senior = 500*800 = 400,000.
Total number of sibling pairs = 30.
P(sibling pair) = 30/400,000 = 3/40,000.
The correct answer is A.
Last edited by GMATGuruNY on Mon Apr 13, 2015 3:11 am, edited 1 time in total.
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Here's another approach:topspin360 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 each school, what is the probability that a sibling pair is selected?
A) 3/40000
B) 3/20000
C) 3/4000
D) 9/400
E) 6/130
OAA
P(selecting a sibling pair) = P(select a business student with a sibling AND select a law student who is that business student's sibling)
= P(select a business student with a sibling) x P(select a law student who is that business student's sibling)
= 30/500 x 1/800
= 30/400,000
= [spoiler]3/40,000[/spoiler]
= A
Note: P(select a business student with a sibling) = 30/500, because 30 of the 500 business students have a sibling in law school.
P(select a law student who is that business student's sibling) = 1/800, because there are 800 law students and only 1 is the sibling of the selected business student.
Cheers,
Brent
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Hi topspin360,
In certain types of probability questions (in which you're dealing with multiple "events"), the result of one event may affect the "math" of the other events.
Here, we're asked to calculate the probability that a pair of siblings is selected (1 from the Business School and the matching sibling from the Law School).
The first step is to calculate the odds of pulling ANY sibling from the Business School:
30/500
So, now we have a sibling from the Business School, but we need to also calculate the probability of choosing the MATCHING sibling from the Law School. There's ONLY ONE matching sibling though (once we choose ANY of the 30 siblings from the Business School, there is just ONE match at the Law School):
1/800
Multiply the results:
(30/500)(1/800) = 3/40,000
Final Answer: A
When tackling these types of questions, it's usually best to work through one calculation at a time and ask yourself if the latter calculations are affected by the first.
GMAT assassins aren't born, they're made,
Rich
In certain types of probability questions (in which you're dealing with multiple "events"), the result of one event may affect the "math" of the other events.
Here, we're asked to calculate the probability that a pair of siblings is selected (1 from the Business School and the matching sibling from the Law School).
The first step is to calculate the odds of pulling ANY sibling from the Business School:
30/500
So, now we have a sibling from the Business School, but we need to also calculate the probability of choosing the MATCHING sibling from the Law School. There's ONLY ONE matching sibling though (once we choose ANY of the 30 siblings from the Business School, there is just ONE match at the Law School):
1/800
Multiply the results:
(30/500)(1/800) = 3/40,000
Final Answer: A
When tackling these types of questions, it's usually best to work through one calculation at a time and ask yourself if the latter calculations are affected by the first.
GMAT assassins aren't born, they're made,
Rich
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Jeff@TargetTestPrep
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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:topspin360 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 each school, what is the probability that a sibling pair is selected?
A) 3/40000
B) 3/20000
C) 3/4000
D) 9/400
E) 6/130
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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Here's another approach:topspin360 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 each school, what is the probability that a sibling pair is selected?
A) 3/40000
B) 3/20000
C) 3/4000
D) 9/400
E) 6/130
OAA
P(selecting a sibling pair) = P(select a business student with a sibling AND select a law student who is that business student's sibling)
= P(select a business student with a sibling) x P(select a law student who is that business student's sibling)
= 30/500 x 1/800
= 30/400,000
= [spoiler]3/40,000[/spoiler]
= A
Note: P(select a business student with a sibling) = 30/500, because 30 of the 500 business students have a sibling in law school.
P(select a law student who is that business student's sibling) = 1/800, because there are 800 law students and only 1 is the sibling of the selected business student.
Cheers,
Brent
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regor60
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To answer this question more directly, you can approach it this way but you 've made a mistake in assuming that picking a student from one school "first" is equally probable with picking a student "first" from the other school.topspin360 wrote:
For the question below, why isn't the answer: (30/500 * 1/800) + (30/800 * 1/500). We can the first student either from business school or law school...
Since there are 1300 students in total, the probability of picking a business student first is 500/1300.
So, you need to multiply (30/500 * 1/800) by 500/1300 from your above question.
Likewise, the probability of picking a law student first is 800/1300, so you need to multiply (30/800*1/500) by this
Combining: (500/1300)*(30/500)*(1/800) + (800/1300)*(30/800)*(1/500)
Notice that the underlined items above are both equal to (30/(800*500), so you can pull this out and rearrange the above:
(30/(500*800))*[(500/1300) + (800/1300)] = 30/(500*800)*[1300/1300] = 30/(500*800) = [spoiler]3/40000, A[/spoiler]
So, this explains the general statement you may have heard that it doesn't matter which order you select
