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:
Probability - Confused with order [#permalink]
14 Feb 2012, 17:45
Problem 1: Source Gmat Club
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]
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.
Re: Probability - Confused with order [#permalink]
14 Feb 2012, 22:19
Expert's post
agarwalmanoj2000 wrote:
Problem 1: Source Gmat Club
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]
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
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...