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A certain junior class has 1,000 students and a certain

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A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?

A. 3/40,000
B. 1/3,600
C. 9/2,000
D. 1/60
E. 1/15

Originally posted by blog on 23 Jan 2008, 12:55.
Last edited by NoHalfMeasures on 26 Dec 2015, 09:18, edited 3 times in total.
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 09 Sep 2010, 21:27
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blog wrote:
A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs , each consisting of 1 junior and 1 senior. If i student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?

A. 3/40,000
B. 1/3,600
C. 9/2,000
D. 1/60
E. 1/15


There are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.

What is the probability of choosing ANY sibling from junior class? \(\frac{60}{1000}\) (as there are 60 of them).

What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be \(\frac{1}{800}\) (as there is only one sibling pair of chosen one).

So the probability of that the 2 students selected will be a sibling pair is: \(\frac{60}{1000}*\frac{1}{800}=\frac{3}{40000}\)

Answer: A.

This problem can be solved in another way:

In how many ways we can choose 1 person from 1000: \(C^1_{1000}=1000\);
In how many ways we can choose 1 person from 800: \(C^1_{800}=800\);
So total # of ways of choosing 1 from 1000 and 1 from 800 is \(C^1_{1000}*C^1_{800}=1000*800\) --> this is total # of outcomes.

Let’s count favorable outcomes: 1 from 60 - \(C^1_{60}=60\);
The pair of the one chosen: \(C^1_1=1\)
So total # of favorable outcomes is \(C^1_{60}*C^1_1=60\)

\(Probability=\frac{# \ of \ favorable \ outcomes}{Total \ # \ of \ outcomes}=\frac{60}{1000*800}=\frac{3}{40000}\).

Answer: A.

Let’s consider another example:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

The same way here:

What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).

So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000

Another way:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.

Favorable outcomes:
1 from 120=120C1=120
The pair of the one chosen=1C2=2
So total favorable outcomes=120C1*1C2=240

Probability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000

Also discussed at: probability-85523.html?hilit=certain%20junior%20class#p641153
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 23 Jan 2008, 13:10
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Answer A

Let's say you're picking out of the Junior class first and the senior class second (although the order doesn't make any difference). There are 1000 juniors and 60 of them have a sibling in the senior class, so you have a Image shot of choosing one of the siblings. Then you move onto the senior class. There are 800 seniors and only one sibling of the person you chose from the junior class. Thus, you have a Image chance of choosing the sibling.

Multiply the two equations together and simplify...and there's your answer.
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 23 Jan 2008, 13:24
Hi,
Great logic. But mez confused that the "1" sibling you've picked from the senior class is from the entire 800 and not from the 60 possible siblings in the senior class.

Does 1/800 imply picking 1 in 60 in this context?
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New post 23 Jan 2008, 13:30
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You have already chosen 1 of the 60 siblings in the junior class. Now that one person is chosen there is only 1 way to choose that persons sibling in the senior class. If you used out of 60 for both classes you would just be finding the probability of finding two people WITH siblings, not necessarily two people who ARE siblings.
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 23 Jan 2008, 14:32
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Hi again.
Thanks for the post.

I do know the stuff you were saying... chk this one -

Total no of pairs available:
1000 * 800 = 800,000

Total no of sibling pairs:
60

So, prob of pickin pair from the lot:
60/800000 = 3/40000

Ans: A :-D
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 17 May 2009, 20:24
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Another way to view this problem:

Probability to get a sibling member on junior class=P1=60/1.000
Probability to get a sibling member on senior class=P2=60/800
Probability to get a sibling pair on all sibling pairs=P3=1/60
Probability that the 2 students selected at will be a sibling pair=P4

P4=P1*P2*P3=(60/1.000)*(60/800)*(1/60)=3/40.000
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 14 Feb 2010, 10:16
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blog wrote:
A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs , each consisting of 1 junior and 1 senior. If i student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?

A. 3/40,000
B. 1/3,600
C. 9/2,000
D. 1/60
E. 1/15


No of ways of choosing 1 sibling pair out of 60 pairs = 60c1
No of ways of choosing 1 student from each class = 1000c1 x 800c1

Therefore probability of having 2 students choosen as a sibling pair = 60c1 / (1000c1 x 800c1) = 60 / (1000 x 800) = 3 / 40000 = A
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 08 Sep 2010, 04:02
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Choosing one pair out of 60 is 60c1
choosing one sibling out of 1000 and one out of 800 is 1000c1*800c1
Probability = Desired/Total
So 60c1/(1000c1*800c1)

A
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 03 Nov 2010, 12:43
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Hi Bunuel,

You have selected first from junior section and then from senior section(first method).
But I have a doubt,its not mentioned anywhere that we have to pick first fro.m junior sec and then from senior sec.
Likewise we have freedom to select first from senior section and then from junior section.
So prob = 60/1000 * 60/800*1/60 + 60/800*60/1000*1/60
= 3/40000 +3/40000
= 6/40000 (Ans)

Please correct me where I am going wrong?
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New post 03 Nov 2010, 13:40
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diptich12 wrote:
Hi Bunuel,

You have selected first from junior section and then from senior section(first method).
But I have a doubt,its not mentioned anywhere that we have to pick first fro.m junior sec and then from senior sec.
Likewise we have freedom to select first from senior section and then from junior section.
So prob = 60/1000 * 60/800*1/60 + 60/800*60/1000*1/60
= 3/40000 +3/40000
= 6/40000 (Ans)

Please correct me where I am going wrong?


Sibling pair (\(a_{junior}\), \(a_{senior}\)) is the same pair as (\(a_{senior}\), \(a_{junior}\)) and with your approach you are counting the probability of selecting each such pair twice.

Sometimes for probability questions it's easy to check whether your approach is right by simplifying the problem. Basically you are saying that the probability is twice as high (instead of A. \(\frac{3}{40000}\), you are saying it's \(\frac{6}{40000}\)).

Consider there is 1 sibling pair, 1 in junior class, with total of 3 students and another in senior class, with total of 2 students. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?

With my approach the answer would be \(\frac{1}{3}*\frac{1}{2}=\frac{1}{6}\) (there are 6 pairs possible and there is only 1 sibling pair). To check whether this answer is correct you can easily list all possible pairs;
With your approach the answer would be: \(\frac{1}{3}*\frac{1}{2}+\frac{1}{2}*\frac{1}{3}=\frac{2}{6}\) , which is not correct. Basically with this approach you are doublecounting the same pair.

Hope it's clear.
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 08 Apr 2011, 03:53
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For those wondering over the order of selection ie. first junior then senior or first senior then junior...
In this case even the total outcomes would double up...
Then you would have total outcomes = 1000C1 * 800C1 + 800C1 * 1000C1
Favourable outcomes = (jr)60C1 * 1C1 + (sr)60C1 * 1C1

Thus p = (60+60)/(1000*800 + 800*1000) = 3/ 40000

This is easy to realize if you take just 3 students each in class with two sibling pairs
Jr = a,b,c Sr = A, B, d

Pick Jr, then Sr => pairs = aA, aB, ad, bA, bB,bd, cA, cB,cd total 9 pairs

Pick Sr, then Jr => pairs = Aa, Ab, Ac, Ba, Bb,Bc, da, db,dc total 9 pairs (just reverse of former 9)

P = 2+2/9+9 = 2/9

Probability is same.
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 07 Dec 2011, 00:07
very good explanation bunnel. +1 for u
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 23 Jul 2013, 19:22
i just wanted to validate this answer the other way around i,e to find the no. of ways at least one sibling or no sibling is present in the chosen 2 members and then subtracting from 1 i.e

Req prob =1- (prob of one sibling chosen from either class) + no sibling chosen from either class
= 1-(prob of one sibling from junior) + (prob of one siblng from senior ) + (no sibling)
= 1-\((\frac{60}{1000}*\frac{799}{800})+(\frac{60}{800}*\frac{999}{1000})+(\frac{740}{800}*\frac{940}{1000})\)


which is coming out to be -ve which obviously is wrong ,, i want to know as to what i am adding extra as a result the answer is -ve
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 23 Jul 2013, 23:04
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adg142000 wrote:
i just wanted to validate this answer the other way around i,e to find the no. of ways at least one sibling or no sibling is present in the chosen 2 members and then subtracting from 1 i.e

Req prob =1- (prob of one sibling chosen from either class) + no sibling chosen from either class
= 1-(prob of one sibling from junior) + (prob of one siblng from senior ) + (no sibling)
= 1-\((\frac{60}{1000}*\frac{799}{800})+(\frac{60}{800}*\frac{999}{1000})+(\frac{740}{800}*\frac{940}{1000})\)


which is coming out to be -ve which obviously is wrong ,, i want to know as to what i am adding extra as a result the answer is -ve


It should be: \(1-(\frac{60}{1000}*\frac{799}{800}+\frac{940}{1000}*1)=\frac{3}{40000}\)

\(\frac{60}{1000}*\frac{799}{800}\) --> p(sibling)*p(any but sibling pair)
\(\frac{940}{1000}*1\) --> p(not sibling)*p(any)

Hope it's clear.
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 30 Sep 2013, 02:35
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blog wrote:
A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs, each consisting of 1 junior and 1 senior. If i student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?

A. 3/40,000
B. 1/3,600
C. 9/2,000
D. 1/60
E. 1/15


Quote:
Responding to a pm: Why doesn't order matter here?


Check out this post: http://www.veritasprep.com/blog/2013/08 ... er-matter/
It discusses a very similar situation.

Also, whether order matter or not depends on how you perceive it. In probability, you calculate P(A) = P(Favorable Outcomes)/P(Total Outcomes)

Where applicable, you can make the order matter in both numerator and denominator or not make it matter in both numerator and denominator. The answer will be the same.

This Question:

Order Matters:
Favorable outcomes = 60*1 + 60*1
Total outcomes = 1000*800 + 800*1000
P(A) = 3/40,000

Order doesn't matter:
Favorable outcomes = 60*1
Total outcomes = 1000*800
P(A) = 3/40,000

Another case: A bag has 4 balls - red, green, blue, pink (Equal probability of selecting each ball)
What is the probability that you pick two balls and they are red and green?

Order Matters:
Favorable outcomes = 2 (RG, GR)
Total outcomes = 4*3
P(A) = 1/6

Order doesn't matter:
Favorable outcomes = 1 (a red and a green)
Total outcomes = 4C2 = 6
P(A) = 1/6

Previous case: 2 red and 3 white balls. What is the probability that you pick two balls such that one is red and other is white?
In how many ways can you pick a red and a white ball? Here the probability of picking each ball is different. So we cannot cannot use our previous method. Here the probability of picking 2 red balls is not the same as that of picking a red and a white. Hence you have to consider the order to find the complete probability of picking a red and a white.
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 04 Dec 2017, 11:53
blog wrote:
A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?

A. 3/40,000
B. 1/3,600
C. 9/2,000
D. 1/60
E. 1/15


The probability of selecting any one sibling from the 60 sibling pairs in the junior class is 60/1000. Once that person is selected, the probability of selecting his or her sibling from the senior class is 1/800; thus, the probability of a selecting a sibling pair is:

60/1000 x 1/800 = 3/50 x 1/800 = 3/40000

Alternatively, the probability of selecting any one sibling from the 60 sibling pairs in the senior class is 60/800. Once that person is selected, the probability of selecting his or her sibling from the junior class is 1/1000; thus, the probability of a selecting a sibling pair is:

60/800 x 1/1000 = 3/40 x 1/1000 = 3/40000

Answer: A
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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 23 Feb 2018, 08:37
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blog wrote:
A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?

A. 3/40,000
B. 1/3,600
C. 9/2,000
D. 1/60
E. 1/15


One option is to apply probability rules.

So, for two siblings to be selected, 2 things must happen: we must select a junior who has a senior sibling AND the senior selected must be the sibling of the selected junior.
We get P(junior with sibling AND selected senior is sibling to selected junior)= P(junior with sibling) x P(selected senior is sibling to selected junior)

P(junior with sibling): there are 1000 juniors and 60 of them have senior siblings. So, P(junior with sibling)=60/1000

P(selected senior is sibling to selected junior): Once the junior has been selected, there is only 1 senior (out of 800 seniors) who is the sibling to the selected junior. So, P(selected senior is sibling to selected junior)= 1/800

So, the probability is (60/1000)x(1/800) = 3/40,000

The answer is A

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Re: A certain junior class has 1,000 students and a certain  [#permalink]

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New post 16 Sep 2018, 15:07
GMATPrepNow wrote:
blog wrote:
A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?

A. 3/40,000
B. 1/3,600
C. 9/2,000
D. 1/60
E. 1/15


\(\left( J \right)\,\,{\text{Junior}}\,\,:\,\,\,1000\,\,{\text{students}}\,\,,\,\,\,{J_1}\,,\,{J_2},\,\, \ldots \,\,,\,\,{J_{60}}\,\,{\text{among}}\,\,{\text{them}}\,\,\,\left( {{\text{the}}\,\,{\text{ones}}\,\,{\text{with}}\,\,{\text{senior}}\,\,{\text{siblings}}} \right)\)

\(\left( S \right)\,\,{\text{Senior}}\,\,:\,\,\,800\,\,{\text{students}}\,\,,\,\,\,{S_1}\,,\,{S_2},\,\, \ldots \,\,,\,\,{S_{60}}\,\,{\text{among}}\,\,{\text{them}}\,\,\,\left( {{\text{the}}\,\,{\text{ones}}\,\,{\text{with}}\,\,{\text{junior}}\,\,{\text{siblings}}} \right)\)

\(\left( {{J_1}\,,\,{S_1}} \right)\,\,;\,\,\left( {{J_2}\,,\,{S_2}} \right)\,\,;\,\, \ldots \,\,;\,\,\left( {{J_{60}}\,,\,{S_{60}}} \right)\,\,\,:\,\,\,{\text{pairs}}\,\,{\text{of}}\,\,{\text{siblings}}\,\)


\(? = P\left( {{\text{pair}}\,{\text{of}}\,{\text{siblings}}\,,\,\,{\text{in}}\,{\text{one}}\,J\,{\text{and}}\,{\text{one}}\,S\,{\text{extraction}}} \right)\)


\({\text{Total}}\,\,:\,\,\,1000 \cdot 800\,\,\,{\text{equiprobables}}\,\,\,\left[ {\left( {{J_m},{S_n}} \right)\,\,,\,\,\,{\text{where}}\,\,1 \leqslant m \leqslant 1000\,\,{\text{and}}\,\,\,1 \leqslant n \leqslant 800} \right]\)

\({\text{Favorable}}\,\,:\,\,60\,\,\,\,\,\left[ {\left( {{J_k},{S_k}} \right)\,\,,\,\,\,{\text{where}}\,\,1 \leqslant k \leqslant 60} \right]\,\,\,\,\)


\(? = \frac{{60}}{{1000 \cdot 800}} = \underleftrightarrow {\frac{{4 \cdot 15}}{{1000 \cdot 4 \cdot 200}}} = \frac{{4 \cdot 3 \cdot 5}}{{1000 \cdot 4 \cdot 5 \cdot 40}} = \frac{3}{{40 \cdot 1000}}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
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