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

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12 Jul 2013, 22:12

eschn3am wrote:

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.

Hi ,

Don't we need to consider the case where we pick the senior class first and then the junior class.

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.

Hi ,

Don't we need to consider the case where we pick the senior class first and then the junior class.

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

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23 Jul 2013, 18:22

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
_________________

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)

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.
_________________

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

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11 Nov 2013, 18:44

Bunuel 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 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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12 Dec 2013, 04:58

jeeteshsingh 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 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

This was What I thought when I solved this question

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

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31 Jan 2014, 13:18

Archit143 wrote:

Can anyone clear my doubt i a m struggling TOtal number of outcome = 1000*800 Sibbling with Juniors = 60 so 60 out of 1000

Sibling with Senior = 60 so 60 out of 800 we have to find probability of 2 sibbling 1 from each 60 C 1= Selection from Senior 60 C 1 = From Junior total fav = 1000*800 Prob ={ 60 C 1 *60 C 1}/1000*800 =9/2000

pls tell me where i am wrong in my approach and what correction needed

For the selection from Junior we would have only 1 and not 60 C 1. This is because since we have chosen someone from the senior class and now the only choice we have is the sibling of the senior student we chose.

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

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07 Feb 2015, 02:51

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

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29 Jul 2015, 23:04

Archit143 wrote:

Can anyone clear my doubt i a m struggling TOtal number of outcome = 1000*800 Sibbling with Juniors = 60 so 60 out of 1000

Sibling with Senior = 60 so 60 out of 800 we have to find probability of 2 sibbling 1 from each 60 C 1= Selection from Senior 60 C 1 = From Junior total fav = 1000*800 Prob ={ 60 C 1 *60 C 1}/1000*800 =9/2000

pls tell me where i am wrong in my approach and what correction needed

Mistake that you are doing is that you considering 60 sibling pairs are present in each junior and senior class. But the question says 60 sibling pairs are present and out of these pair one is present in junior and its corresponding sibling in senior. That means 30 in junior and its opposite pair 30 in senior.
_________________

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

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15 Apr 2016, 23:45

another easy understanding total ways to select the= 1000 X 800 (as each of 1000 juniors can be selected with 800 options) now selecting a sibling = 60 ways X 1 way (as there are 60 siblings so we have 60 ways of selecting on student from junior group and as soon as we select one we are left with only one way to select from senior group as that person has only one sibling in the senior group )

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

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06 Nov 2016, 22:21

Bunuel 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 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

There are 60 sibling pairs. Probability of Selecting One Junior AND their corresponding Senior sibling OR another Junior and their corresponding sibling and so on: Probability of Selecting John is: 1/1000 Probability of Selecting John's brother: 1/800 Number of Pairs: 60

So,

1/1000 x 1/800 x 60 = 30/40000 ways to select any one particular Junior and their corresponding Senior Sibling

[ With the breakdown being: P(John and his sibling) + P (Mary and her sibling) + P(Tom and his sibling) + ..... = (1/1000 x 1/800) + (1/1000 x 1/800) + (1/1000 x 1/800) + ..... and so on, until you add the 60 different possibilities ]

gmatclubot

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