Leap Year gets no love

We know that out of 4 years , 1 is a leap year. So, probability of one person to be born in a leap year is $\frac { 1 }{ 4 }$ and probability of not to be born in a leap year is $\frac { 3 }{ 4 }$ .

So, if there is one person, probability of him to be born in leap year is $\frac { 1 }{ 4 }$ x 100% = 25%

If there are two person, probability of both not to be born in leap year is

$\frac { 3 }{ 4 } \times \frac { 3 }{ 4 } =\frac { 9 }{ 16 } \times 100$ %=56.25% . So, probability of at least one of them to be born in a leap year is (100-56.25)%=43.75% .

Similarly , if there are three person, probability of all the three not to be born in leap year is $\frac { 3 }{ 4 } \times \frac { 3 }{ 4 } \times \frac { 3 }{ 4 } =\frac { 27 }{ 64 } \times 100$ %=42.19% . So, probability of at least one to be born in a leap year is (100-42.19)%=57.81% which is greater than 50% . Hence, we must have at least 3 people to ensure that probability of at least one of them to be born in a leap year is greater than 50% .

For the sake of precision, there are 366 days in a leap year and 365 days in a non-leap year. So, the probability that a person is born in a leap year is $\frac{366}{366+3(365)}$ considering a 4-year cycle, which is approximately 25.051%.

So the probability that a person is not born in a leap year is about 100% - 25.051% =74.949%. Let $n$ be the number of people (we are considering). P(at least 1 person is born in a leap year) = 1 - P(no one was born in a leap year) $=1-(0.74949)^n$ .

Thus, $1-(0.74949)^n> 0.5$ Solving, the least integer value of n is 3.

Think of it as a 4 sided die with numbers 1,2,3,4 and 1 person =1 roll.

The probability of getting a 1 is the same as getting a leap year. After rolling, there is a 1/4 probability of getting a 1, after 2 rolls your chances are 50%.

Note that the question asks for >50%, not $\geq 50\%$

Therefore we must have 3 rolls to have more more than a 50℅ chance.