Problem Solving

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?

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?

Is that the original wording of the question? "randomly assembled people"? What's the source? What are the answer choices?

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%.

exactly answer is 3