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

It is currently 24 Aug 2016, 11:25
GMAT Club Tests

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

Good explanation needed for this one Jane has to pick 3

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Expert Post
Founder
Founder
User avatar
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 13981
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3450

Kudos [?]: 20545 [0], given: 4339

GMAT ToolKit User Premium Member CAT Tests
Good explanation needed for this one Jane has to pick 3 [#permalink]

Show Tags

New post 28 Jul 2003, 19:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Good explanation needed for this one

Jane has to pick 3 subjects out of 6: A, B, C, D, E, or F. If she has already chosen E, what is the probability that she will choose B also?
тАв 0.2
тАв 0.25
тАв 0.4
тАв 0.8
тАв 0.96
GMAT Instructor
User avatar
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 18

Kudos [?]: 167 [0], given: 0

Re: Probability 2 [#permalink]

Show Tags

New post 28 Jul 2003, 20:59
bb wrote:
Good explanation needed for this one

Jane has to pick 3 subjects out of 6: A, B, C, D, E, or F. If she has already chosen E, what is the probability that she will choose B also?
тАв 0.2
тАв 0.25
тАв 0.4
тАв 0.8
тАв 0.96


Here are 3 ways of looking at this:
(1)
WE are already given that E is chosen, so the problem is simply what is the prob that she will pick B if she chooses 2 subjects.

This is the probability that she will either choose the subject as the first pick OR the subject as the second pick.

Pr(b 1st) = 1/5
Pr(b 2nd) = Pr(b not 1st AND b 2nd) = 4/5 * 1/4 = 1/5

These 2 are mutually exclusive, hence the probability is 1/5 + 1/5 = 2/5 = 0.4

Method 2:
The probability that she will NOT leave B as one of her 3 unpicked choice is exactly the same problem.
Pr(B is NOT one of 3 unpicked choices) = 4/5 * 3/4 * 3/2 = 2/5 = 0.4

Method 3:
Pr (b picked in 2 choices) = 1 - Pr(B not picked of her two choices)
= 1 - (4/5 * 3/4) = 1 - 3/5 = 2/5 = 0.4.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
Manager
User avatar
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 2

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

Re: Probability 2 [#permalink]

Show Tags

New post 29 Jul 2003, 06:54
bb wrote:
Good explanation needed for this one

Jane has to pick 3 subjects out of 6: A, B, C, D, E, or F. If she has already chosen E, what is the probability that she will choose B also?
тАв 0.2
тАв 0.25
тАв 0.4
тАв 0.8
тАв 0.96


Another way of looking at this problem:

E is already chosen. Now B can be chosen in 2nd choice or third choice.

Number of ways B can come in second choice = 1C1*4C1 = 4
Number of ways B can come in third choice = 4C1*1C1 = 4

Total number of ways in which two subjects can be chosen, one by one, out of the remaining five = 5C1*4C1 = 20

Therefore, the probability that B will be chosen = 8/20 = 0.4
SVP
SVP
User avatar
Joined: 30 Oct 2003
Posts: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 81 [0], given: 0

 [#permalink]

Show Tags

New post 23 Jan 2004, 08:19
A different approach. Assume he has chosen E and B then only one other subject has to be chosen in 4C1 ways
Total ways to to choose 2 subjects from 5 is 5C2 = 10 ways

P = 4C1/5C2 = 0.4
  [#permalink] 23 Jan 2004, 08:19
Display posts from previous: Sort by

Good explanation needed for this one Jane has to pick 3

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| 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®.