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

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23 Jan 2008, 12:55

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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?

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

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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 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 chance of choosing the sibling.

Multiply the two equations together and simplify...and there's your answer.

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

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

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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

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

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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.

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

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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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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.

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?

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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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)

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

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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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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)

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

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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.

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

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27 Sep 2009, 02:45

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

Soln: Total number of ways of choosing one student from each group is = 800 * 1000 Number of cases in which a sibling pair will be got is = 60

Thus the probability that the 2 students selected will be a sibling pair is = 60 /(800 * 1000) = 3/40,000

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

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08 Sep 2010, 04:02

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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05 Dec 2011, 12:09

(60c1*1c1)/(800c1*1000c1)=3/40,000 I used to struggle with this sort of problem. I followed gmatclub math book (combinatorics). i guess i am improving.

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

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12 Dec 2011, 12:08

P=(probability of selecting one of 60 people from 1000 junior)*(probability of selecting one of 60 people from 800 senior)*(probability of matching these 2 people as a sibling)

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

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12 Dec 2011, 12:14

A lot of math going on up there above this post.

Way I looked at it was you have to pick one of each. If you pick a senior, you have a 60 in 1,000 shot of picking a sibling. There is a 1 in 800 chance you pick his/her pair once you pick again.

So 60/1000 times 1/800 = 60/800,000, which reduces to 3/40,000

gmatclubot

Re: A certain junior class has 1,000 students and a certain
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12 Dec 2011, 12:14

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