Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A certain junior class has 1000 students and a certain [#permalink]

Show Tags

13 Aug 2010, 02:15

3

This post received KUDOS

7

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

70% (01:51) correct
30% (01:07) wrong based on 212 sessions

HideShow timer Statistics

A certain junior class has 1000 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 will be a sibling pair?

A. 3/40000 B. 1/3600 C. 9/2000 D. 1/60 E. 1/15

My explanation is:

Total 60 students are siblings, out of which 30 are from Junior class and 30 are from senior class. Hence prob of selecting 1 student from senior who is a sibling is 30C1/800C1, similarly, selecting one student from Junior who is a sibling is 30C1/1000C1. Since selecting 2 ppl from 2 sets, the events are independent, total probability is : 30/800+ 30/1000. Simplifying, I get 1/15.

First of all we have 60 siblings pairs, so there are 60 siblings in junior class and 60 siblings in senior class.

Next: the question ask "what is the probability that the 2 students selected will be a sibling pair", so the probability that they will be siblings of each other.

Back to the question:

A certain junior class has 1000 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 will be a sibling pair?

A. 3/40000 B. 1/3600 C. 9/2000 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 \(C^1_{60}*C^1_1=60\)

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

217) A certain junior class has 1,000 students and a certain senior class has 800 students.among these students, there are 60 sibling 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 will be a sibling pair.

a) 3/40,0000 b) 1/3600 c) 9/2000 d) 1/60 e) 1/15

Probability to select one senior student who is sibling is 60/800 Probability to select one junior student who is sibling is 60/1000

As both the events are independent and should happen we ill multiply the two probabilities 60/800 * 60/1000 = 9/2000

Pls tell me the error in my solution
_________________

The proof of understanding is the ability to explain it.

217) A certain junior class has 1,000 students and a certain senior class has 800 students.among these students, there are 60 sibling 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 will be a sibling pair.

a) 3/40,0000 b) 1/3600 c) 9/2000 d) 1/60 e) 1/15

Probability to select one senior student who is sibling is 60/800 Probability to select one junior student who is sibling is 60/1000

As both the events are independent and should happen we ill multiply the two probabilities 60/800 * 60/1000 = 9/2000

Pls tell me the error in my solution

Please ask if anything remains unclear.
_________________

217) A certain junior class has 1,000 students and a certain senior class has 800 students.among these students, there are 60 sibling 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 will be a sibling pair.

a) 3/40,0000 b) 1/3600 c) 9/2000 d) 1/60 e) 1/15

Probability to select one senior student who is sibling is 60/800 Probability to select one junior student who is sibling is 60/1000

As both the events are independent and should happen we ill multiply the two probabilities 60/800 * 60/1000 = 9/2000

Pls tell me the error in my solution

If probability of selecting one senior student is 60/800 Probability of selecting the matching pair from the junior students becomes 1/1000

Think it like this; You have successfully chosen 1 sibling of sibling pairs from the senior students. Now; when you start choosing from the juniors; you just have 1 favorable outcome. Because; out of these 1000 students, there is only 1, just ONE student who is the paired sibling of the student you earlier chose from the senior students. Got it?

So; the total probability becomes = 60/800*1/1000 = 3/40000.

Ans: "A"

I believe there was a similar post yesterday. Also; this particular question is also discussed elsewhere. Guess this post is going to be short lived.
_________________

A certain junior class has 1000 students and a certain senio [#permalink]

Show Tags

13 Apr 2012, 01:32

A certain junior class has 1000 students and a certain senior class has 800 students.Among these students there are 60 sibling 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 will be a sibling pair ?

A certain junior class has 1000 students and a certain senior class has 800 students.Among these students there are 60 sibling 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 will be a sibling pair ?

Re: A certain junior class has 1000 students and a certain [#permalink]

Show Tags

17 Apr 2012, 22:19

Anyone please clear my doubt. I have a similar ques A bag has 6 red marbles and 4 marbles. What are the chances of pulling out a red and blue marble. Method 1: # of ways of picking 1 red and 1 blue marble = (6c1)(4c1) = 6 x 4 = 24; # of ways of picking 2 marbles in general = 10c2 = 45. therefore, probability = 24/45 Method 2: Total prob = prob ( R then B) + P (B then R) 6 ways for red and 4 ways for blue = 24 total ways = 10*9 Prob ( R then B) = 24/90

||ly Prob ( B then R) = 24/90 Therfore total = 24/45

Re: A certain junior class has 1000 students and a certain [#permalink]

Show Tags

07 Jul 2014, 20:17

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: A certain junior class has 1000 students and a certain [#permalink]

Show Tags

10 May 2015, 22:30

Bunuel wrote:

First of all we have 60 siblings pairs, so there are 60 siblings in junior class and 60 siblings in senior class.

Next: the question ask "what is the probability that the 2 students selected will be a sibling pair", so the probability that they will be siblings of each other.

Back to the question:

A certain junior class has 1000 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 will be a sibling pair?

A. 3/40000 B. 1/3600 C. 9/2000 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 \(C^1_{60}*C^1_1=60\)

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

Answer: A.

Hope it helps.

Hi Am not good at probability plz clarify

Questions says 1 student is selected from each class shouldn't it be like>>

\(Probability=\frac{60}{1000} * \frac{1}{800} OR \frac{60}{800} * \frac{1}{1000} = \frac{3}{20000}\). _________________

First of all we have 60 siblings pairs, so there are 60 siblings in junior class and 60 siblings in senior class.

Next: the question ask "what is the probability that the 2 students selected will be a sibling pair", so the probability that they will be siblings of each other.

Back to the question:

A certain junior class has 1000 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 will be a sibling pair?

A. 3/40000 B. 1/3600 C. 9/2000 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 \(C^1_{60}*C^1_1=60\)

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

Answer: A.

Hope it helps.

Hi Am not good at probability plz clarify

Questions says 1 student is selected from each class shouldn't it be like>>

\(Probability=\frac{60}{1000} * \frac{1}{800} OR \frac{60}{800} * \frac{1}{1000} = \frac{3}{20000}\).

The order of selecting the sibling does not matter here. Let me explain you why your probability equation is not correct. Assume a pair of siblings A & B where A is in the senior class & B is in the junior class. If you select A from the senior class first, you can only select B from the junior class to make it a sibling pair.

Alternatively, if you select B from the junior class first, you can only select A from the senior class to make it a sibling pair. Thus, in both the cases we have the same pair of siblings as our final selection . Hence the order of selection of siblings does not matter which is what your probability equation is intending to convey.

Re: A certain junior class has 1000 students and a certain [#permalink]

Show Tags

10 May 2015, 23:30

EgmatQuantExpert wrote:

dpo28 wrote:

Bunuel wrote:

First of all we have 60 siblings pairs, so there are 60 siblings in junior class and 60 siblings in senior class.

Next: the question ask "what is the probability that the 2 students selected will be a sibling pair", so the probability that they will be siblings of each other.

Back to the question:

A certain junior class has 1000 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 will be a sibling pair?

A. 3/40000 B. 1/3600 C. 9/2000 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 \(C^1_{60}*C^1_1=60\)

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

Answer: A.

Hope it helps.

Hi Am not good at probability plz clarify

Questions says 1 student is selected from each class shouldn't it be like>>

\(Probability=\frac{60}{1000} * \frac{1}{800} OR \frac{60}{800} * \frac{1}{1000} = \frac{3}{20000}\).

The order of selecting the sibling does not matter here. Let me explain you why your probability equation is not correct. Assume a pair of siblings A & B where A is in the senior class & B is in the junior class. If you select A from the senior class first, you can only select B from the junior class to make it a sibling pair.

Alternatively, if you select B from the junior class first, you can only select A from the senior class to make it a sibling pair. Thus, in both the cases we have the same pair of siblings as our final selection . Hence the order of selection of siblings does not matter which is what your probability equation is intending to convey.

Hope its clear!

Regards Harsh

Hi Harsh I think my doubt is clear now in both the cases the pair will be the same thnx 4 the reply
_________________

Re: A certain junior class has 1000 students and a certain [#permalink]

Show Tags

26 Jun 2016, 10:05

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...