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