knight247 wrote:Lets assume there are N people
Out of these N ppl, the probability that any one of them was born in a leap year is 1/4
Probability that the same person is not born in a leap year is 1-1/4=3/4
So the probability that all N ppl are not born in a leap year is (3/4)^N
Now,
1-(3/4)^N is the probability that atleast one of them is born in a leap year.
It needs to be greater than 1/2 so we use the inequality
1-(3/4)^N>1/2
1/2>(3/4)^N
(3/4)^N<1/2
Now try plugging in different values of N from the answer options
(A)if N=1 then
3/4>1/2 so NO
(B)N=2
9/16>1/2 so NO
(C)N=3
27/64<1/2 so YES
(D)N=4
81/256<1/2 so YES
(E)N=5
243/1024<1/2 so YES
I guess the question should have asked what is the minimum number of ppl. The question doesn't seem to be properly worded. Anyway, if it is the minimum number of ppl then the answer is C
Hishankar.ashwin wrote: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
This is a common problem.shekhar.kataria wrote: I see many questions which i feel are beyond the scope of GMAT.
Agreed - the probability component of this question is within the scope of the GMAT.shankar.ashwin wrote:I got it in a practice test, but as Brent says maybe Leap Year questions may not appear in the GMAT, but that said the probability part of this question is pretty straightforward and IMO can be expected in the GMAT especially on high difficulty quants questions. I think experts would agree too.
