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A certain business school has 500 students, and the law school at the

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Carcass
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A certain business school has 500 students, and the law school at the [#permalink] New post   22 Oct 2021, 06:55
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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. \(\frac{6}{130}\)
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Re: A certain business school has 500 students, and the law school at the [#permalink] New post   22 Oct 2021, 07:16
Carcass 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. \(\frac{6}{130}\)



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
= 3/40,000

Answer: 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.

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Brent
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Re: A certain business school has 500 students, and the law school at the [#permalink] New post   29 Jul 2023, 05:29
You counted: (30/500)*(1/800)
But what about: (30/800)*(1/500)?

(30/500)*(1/800) + (30/800)*(1/500) = 60/400000 = 3/20000

Answer: B

Why is it wrong?


GreenlightTestPrep wrote:
Carcass 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. \(\frac{6}{130}\)



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
= 3/40,000

Answer: 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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Re: A certain business school has 500 students, and the law school at the [#permalink] New post   29 Jul 2023, 05:35
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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 school at the [#permalink] New post   29 Jul 2023, 05:35
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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(b-l) = 1/800
P9l-b) = 1/500

3rd, you need to multiply resepectively ("AND" - for example - both P(l) and P(b-l) 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 school at the [#permalink] New post   29 Jul 2023, 06:53
Cab you please explain the first part of this approach?
P(l) = 15/500
P(b) = 15/800

Why did you put 15 students in each school?
It talks about 30 pairs of siblings, so there are 60 students.
So each school has 30 students who have siblings in the other school, no?

Thank you for the help, I just confused by the 15 number.

Carcass wrote:
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(b-l) = 1/800
P9l-b) = 1/500

3rd, you need to multiply resepectively ("AND" - for example - both P(l) and P(b-l) 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 school at the [#permalink] New post   29 Jul 2023, 07:12
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Because they treat a couple as one

or you can think this way

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 school at the [#permalink]
29 Jul 2023, 07:12
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