BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Leap year probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 407
Joined: Tue Jan 25, 2011 9:19 am
Thanked: 25 times
Followed by:7 members

Leap year probability

by Ozlemg » Sun Aug 21, 2011 4:52 am
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

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

OA : c
Source : gmatclub.forum
The more you suffer before the test, the less you will do so in the test! :)
Top
Source: — Problem Solving |
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Sun Aug 21, 2011 7:02 am
Hi,
Let's say there are 'n' people.
Probability that a person was born in a non-leap year is 3/4
So, probability that all n people are born in non-leap year is (3/4)^n.
So, probability that at least one of them is born in a leap year is 1 - (3/4)^n
We need to find the minimum value of 'n' for which 1-(3/4)^n > 1/2 => (3/4)^n < 1/2
For n=1, 3/4 > 1/2
For n=2, 9/16 > 1/2
For n=3, 27/64 < 1/2

Hence, C
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Senior | Next Rank: 100 Posts
Posts: 96
Joined: Wed May 20, 2009 8:46 pm
Thanked: 1 times

by jonathan123456 » Sun Aug 21, 2011 12:52 pm
Frankenstein wrote:Hi,
Let's say there are 'n' people.
Probability that a person was born in a non-leap year is 3/4
So, probability that all n people are born in non-leap year is (3/4)^n.
So, probability that at least one of them is born in a leap year is 1 - (3/4)^n
We need to find the minimum value of 'n' for which 1-(3/4)^n > 1/2 => (3/4)^n < 1/2
For n=1, 3/4 > 1/2
For n=2, 9/16 > 1/2
For n=3, 27/64 < 1/2

Hence, C
Can you show how we can get the same answer by using leap year prob i.e 4
Atleast 1 + Atleast 2 + Atleast3 ?

Thanks
Never say Die!
Top
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Mon Aug 22, 2011 12:28 am
jonathan123456 wrote: Can you show how we can get the same answer by using leap year prob i.e 4
Atleast 1 + Atleast 2 + Atleast3 ?

Thanks
Hi,
Sorry I am unable to understand your post. Could you make it better so that I can answer your query.
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Post new topic Post Reply
• Page 1 of 1