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:

Set consists of all prime integers less than 10. If a number is selected from set at random and then another number, not necessarily different, is selected from set at random, what is the probability that the sum of these numbers is odd?

Answer: Set S =(2,3,5,7) . The question "what is the probability that the sum of these numbers is odd?" is equivalent to the question "what is the probability that one of these numbers is 2 while the other is not?".

P(the sum is odd) = (P(the first number is 2) * P (the second number is not 2)) + (P(the first number is NOT 2) * P (the second number is 2)) = (¼ * 3/4 ) + (¾ * ¼) = 3/8

P(the sum is odd) = (P(the first number is 2) * P (the second number is not 2)) + (P(the first number is NOT 2) * P (the second number is 2)) = (¼ * 3/4 ) + (¾ * ¼) = 3/8

Problem 2: Source: Do not remember.

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.

Answer: First the probability of selecting one of the siblings from the business school is 30/500. After we have made that selection, the probability of selecting appropriate sibling from the law school is 1/800. Therefore the probability that we select a sibling pair is 30/500 * 1/800 = 3/40000

My question:

In problem 1, probability of selecting the first number 2 and probability of selecting second number 2 is found i.e. P(the sum is odd) = (P(the first number is 2) * P (the second number is not 2)) + (P(the first number is NOT 2) * P (the second number is 2)) = (¼ * 3/4 ) + (¾ * ¼) = 3/8

But, in problem 2, only probability of selecting first sibling from business school is found i.e. (P(the first sibling from business school) * P (the second sibling from law school)) = 30/500 * 1/800 = 3/40000

Why two cases are not considered, similar to problem 1 i.e.

P(selecting sibling) = (P(the first sibling from business school) * P (the second sibling from law school)) + (P(the first sibling from law school) * P (the second sibling from business school)) = (30/500 * 1/800) + (1/800 * 30/500) = 3/20000

Re: Probability - Confused with order [#permalink]

Show Tags

14 Feb 2012, 19:33

Because there are 2 ways of choosing the desire outcome in the first problem. In the first Q you want one two. There is a 3/4 chance to not get 2 and a 1/4 chance to get two. If you don't get a 2 on the first row, you still have a probability to get 2 in the second row and vice versa for if you got two first. Since there is a 1/4 probability to get 2 in the first row and a 3/4 probability to get not 2 in the second row. So you add the total probability of the desire outcome.

For question 2. If you don't pick a twin on the first pick, then it doesn't matter what you pick in the second pick. There isn't any way to get the two different desirable outcomes. Unlike in the first example, if you don't get 2 there is another option to achieve the desire outcome. If you don't get a twin you can't get the desirable outcome period. So there is only one way to get the twins.

Set consists of all prime integers less than 10. If a number is selected from set at random and then another number, not necessarily different, is selected from set at random, what is the probability that the sum of these numbers is odd?

Answer: Set S =(2,3,5,7) . The question "what is the probability that the sum of these numbers is odd?" is equivalent to the question "what is the probability that one of these numbers is 2 while the other is not?".

P(the sum is odd) = (P(the first number is 2) * P (the second number is not 2)) + (P(the first number is NOT 2) * P (the second number is 2)) = (¼ * 3/4 ) + (¾ * ¼) = 3/8

P(the sum is odd) = (P(the first number is 2) * P (the second number is not 2)) + (P(the first number is NOT 2) * P (the second number is 2)) = (¼ * 3/4 ) + (¾ * ¼) = 3/8

Problem 2: Source: Do not remember.

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.

Answer: First the probability of selecting one of the siblings from the business school is 30/500. After we have made that selection, the probability of selecting appropriate sibling from the law school is 1/800. Therefore the probability that we select a sibling pair is 30/500 * 1/800 = 3/40000

My question:

In problem 1, probability of selecting the first number 2 and probability of selecting second number 2 is found i.e. P(the sum is odd) = (P(the first number is 2) * P (the second number is not 2)) + (P(the first number is NOT 2) * P (the second number is 2)) = (¼ * 3/4 ) + (¾ * ¼) = 3/8

But, in problem 2, only probability of selecting first sibling from business school is found i.e. (P(the first sibling from business school) * P (the second sibling from law school)) = 30/500 * 1/800 = 3/40000

Why two cases are not considered, similar to problem 1 i.e.

P(selecting sibling) = (P(the first sibling from business school) * P (the second sibling from law school)) + (P(the first sibling from law school) * P (the second sibling from business school)) = (30/500 * 1/800) + (1/800 * 30/500) = 3/20000

Now, for the 1st question we are interested in getting the odd sum: {Odd, Even} and {Even, Odd} are two different outcomes of favorable scenario and we should count both cases. For example {2, 3} is different case from {3, 2}: we can get the sum of 5 in two different ways, first chosen number is 2 and the second chosen number is 3 or first chosen number is 3 and the second chosen number is 2.

For the 2nd question we are interested in selecting a sibling pair. Suppose there are sibling pair A and A': {A, A'} and {A', A} are the same pair and no need to count them twice (favorable scenario is {A, A'}).

Re: Probability - Confused with order [#permalink]

Show Tags

15 Feb 2012, 22:22

Dear Bunnel,

Thanks a lot for your reply and great links. I read them all.

In question 1, both {2, 3} and {3, 2} gives odd sum, and in question 2, both {A, A'} and {A', A} gives sibling pair. Why {2, 3} and {3, 2} are different but {A, A'} and {A', A} are same?

How to know in which case I need to count twice and in which case I need to count once? Is there some standard or rule?

Please guide.

gmatclubot

Re: Probability - Confused with order
[#permalink]
15 Feb 2012, 22:22

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...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...