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Leap year probability

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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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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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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
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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!

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