Find all School-related info fast with the new School-Specific MBA Forum

It is currently 20 Oct 2014, 05:13

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Probability - Confused with order

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 30 May 2011
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 2

Probability - Confused with order [#permalink] New post 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

Please guide.
Kaplan Promo CodeKnewton GMAT Discount CodesVeritas Prep GMAT Discount Codes
Manager
Manager
avatar
Joined: 31 Jan 2012
Posts: 74
Followers: 1

Kudos [?]: 17 [0], given: 2

Re: Probability - Confused with order [#permalink] New post 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.
Intern
Intern
avatar
Joined: 30 May 2011
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 2

Re: Probability - Confused with order [#permalink] New post 14 Feb 2012, 21:06
Hi kys123,

Thanks for your reply. I have below queries.

1)
Similar to problem 1, there are two ways to pick siblings in problem 2 i.e.

First way is picking a sibling from business school and then picking a sibling from law school.

Second way is picking a sibling from business school and then picking a sibling from law school.

So why are we considering only one way in problem 2?

2) How to know for which question should we consider one or two case?

Thanks in advance.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23331
Followers: 3599

Kudos [?]: 28598 [0], given: 2803

Re: Probability - Confused with order [#permalink] New post 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

Please guide.


First of all check Probability chapter of Math Book: gmat-math-book-87417.html

Next, see the discussions of the 1st questions:
set-s-consists-of-all-prime-integers-less-than-10-if-a-82170.html
m10-q-100026.html

Discussions of the 2nd question:
probability-85523.html
a-certain-junior-class-has-1000-students-and-a-certain-99046.html
in-a-room-filled-with-7-people-4-people-have-exactly-87550.html
a-certain-junior-class-has-1000-students-and-a-certain-87909.html

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'}).

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 30 May 2011
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 2

Re: Probability - Confused with order [#permalink] New post 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.
Re: Probability - Confused with order   [#permalink] 15 Feb 2012, 22:22
    Similar topics Author Replies Last post
Similar
Topics:
1 Experts publish their posts in the topic Confused with probablity kairoshan 13 19 Nov 2009, 05:58
Confusing Probability Question aliassad 2 24 Aug 2010, 08:30
ordering ritula 2 03 Feb 2009, 03:25
This probably is confusing to me. The figure shows the top twinkle9600 11 31 Dec 2005, 14:04
Experts publish their posts in the topic Confused just4fun 3 06 Dec 2004, 02:47
Display posts from previous: Sort by

Probability - Confused with order

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.