GMAT Probability - Selection from a group

How many randomly assembled people do u need to have a better than 50% probability that at least 1 of them was born in a leap year?

3
9
27
54
81



Let's get a formula to work with. If that doesn't yield an answer, then we can plug in answer choices since they're spread far apart.


The probability that one of them was born in a leap year needs to be greater than 50%
P(A) = 1 - P(A')


Manipulate this algebraicly

1 - Probability that none of them was born in a leap year > 1/2
Prob[none] - 1 < -1/2
Prob[none] < 1/2

Let's say there are n people.

Prob[none] = (3/4)^n
(3/4)^n < 1/2
3^n < (4^n)/2
3^n < (2^2n)/2
3^n < 2^(2n-1)

Plug in.
If n = 3, 3^3 < 2^5 checks out
(You can try other answer choices to see they do not work out to viable answers)

A.

Jeremiah LaBrash is a programmer and CIO for a CCaaS telecom company based in New York, NY. If you have math or verbal questions you’re having difficulty with and would like Jeremiah LaBrash to solve them and parse them into understandable parts, please leave a comment below or mail jr@thelevel11.com