Leap year probability

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Ozlemg: Master | Next Rank: 500 Posts; Posts: 407; Joined: Tue Jan 25, 2011 9:19 amFollowed by: 7 members; Thanked: 25 times

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

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Source: Beat The GMAT — Problem Solving | Sun Aug 21, 2011 4:52 am

Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

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

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jonathan123456: 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!

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Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

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

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