It is currently 18 Oct 2017, 08:35

Live Now:

GMAT Verbal Live on YouTube: Join Now!


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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

How many randomly assembled people are needed to have a

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
S
Joined: 17 May 2015
Posts: 38

Kudos [?]: 7 [0], given: 166

Premium Member
Re: How many randomly assembled people are needed to have a [#permalink]

Show Tags

New post 05 May 2016, 05:50
Why is intersection of two independent event multiplied?


VeritasPrepKarishma wrote:
reto wrote:
What is the minimum number of randomly chosen people needed in order to have a better-than-50% chance that at least one of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5


Keeping it simple:

P (A person was born in leap year) = 1/4
This is less than 50% - not the answer

P (A or B was born in a leap year) = P (A born in leap year) + P (B born in leap year) - P (Both born in leap year)


The two events are independent. Probability that A was born is a leap year is independent of the probability that B was born in a leap year.
So P(Both) = P(A)*P(B)

P (A or B was born in a leap year) = 1/4 + 1/4 - (1/4)*(1/4) = 1/2 - 1/16

This is less than 50% but close - not the answer

So when you pick, three people, the probability of someone born in a leap year will be higher than 50%.

If you want to calculate it, you can do it as given below:

P(A or B or C was born in a leap year) = 1/4 + 1/4 + 1/4 - 1/16 - 1/16 - 1/16 + 1/64 = 1/2 + 1/16 + 1/64 (same as the sets concept)

This is definitely more than 50%.

Answer (C)

Kudos [?]: 7 [0], given: 166

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 16708

Kudos [?]: 273 [0], given: 0

Premium Member
Re: How many randomly assembled people are needed to have a [#permalink]

Show Tags

New post 16 Aug 2017, 06:54
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 273 [0], given: 0

Re: How many randomly assembled people are needed to have a   [#permalink] 16 Aug 2017, 06:54

Go to page   Previous    1   2   [ 22 posts ] 

Display posts from previous: Sort by

How many randomly assembled people are needed to have a

  new topic 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 | Emoji artwork provided by EmojiOne

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®.