Explanation

(d)
Required probability
$= P($when two persons have some birthday)
$+ P($when three persons have same birthday)
$+ P($when four persons have same birthday)
$= \ ^{4}C_{2} \times \frac{1}{365} \times \left( \frac{364}{365} \right)^{2} + \ ^{4}C_{3}\frac{1}{(365)^{2}} \times \frac{364}{365} + \ ^{4}C_{4}\left( \frac{1}{365} \right)^{3}$
$= \frac{6(364)^{2} + 4 \times 364 + 1}{(365)^{3}}$
$= \frac{794976 + 1456 + 1}{(365)^{3}}$
$= \frac{796433}{48627125} = 0.016$