Leap year!

Since a leap year occurs every $4$ years, (there are exceptions but we'll keep it simple here), the probability that a random person is not born on a leap year is $\frac{3}{4}$ . Thus the probability that $n$ randomly chosen people are all born on non-leap years is $(\frac{3}{4})^{n}$ .

The probability that at least one of $n$ randomly chosen people is born on a leap year is then $1 - (\frac{3}{4})^{n}$ . So what we need to find is the least integer $n$ such that

$(1 - (\frac{3}{4})^{n}) \ge 0.5 \Longrightarrow (\frac{3}{4})^n \le 0.5$ .

We could use logarithms at this point, but it is faster to just plug in integers $1, 2, 3$ to see that $\boxed{3}$ is the least such integer $n$ .

Comment: If we were to use logarithms, then we would need to find the least integer $n$ such that

$n \ge \frac{\ln(\frac{1}{2})}{\ln(\frac{3}{4})} \Longrightarrow n \ge 2.41 \Longrightarrow n = 3$ .