Is that the original wording of the question? "randomly assembled people"? What's the source? What are the answer choices?How many randomly assembled people do u need to have a better than 50% prob. that at least 1 of them was born in a leap year?
With these questions yet to be answered, my solution goes as follows:
p(leap year) = roughly 0.25 (roughly because not every 4th year is a leap year, I guess you can check special leap year rules on wikipedia or elsewhere on the internet, but for the sake of simplicity, let's assume that every 4th year is a leap year)
p(no leap year) = 1 - p(leap year) = 0.75 (again roughly)
For at least 1 person to be born in a leap year out of an unknown number of people with a greater than 50% probability , there has to be a probability smaller than 50% for 0 people to be born in a leap year.
For 1 person:
p(noone born in LY) = 0.75
For 2 people
p(noone born in LY) = 0.75^2 > 0.5
For 3 people
p(noone born in LY) = 0.75^3 < 0.5
Bingo! When you have 3 "randomly assembled" people the probability that at least 1 of them was born in a leap year is bigger than 50%.
