5. A bag contains 30 balls that are numbered $1,2, \ldots, 30$. Two balls are randomly chosen from the bag. Find the probability that the sum of the two numbers is divisible by 3 .
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{7}{29}$
(D) $\frac{9}{29}$
(E) $\frac{11}{87}$
5. Answer: (B)
Let $S_{1}=\{1,4,7, \ldots, 28\}, S_{2}=\{2,5,8, \ldots, 29\}, S_{3}=\{3,6,9, \ldots, 30\}$. Note that each of the three sets has 10 elements. Now for any two distinct numbers $a$ and $b$ in $\{1,2,3, \ldots, 28,29,30\}$, their sum $a+b$ is divisible by 3 if and only if one of the following conditions holds:
(i) the two numbers $a$ and $b$ belongs to $S_{1}$ and $S_{2}$ respectively;
(ii) both numbers $a$ and $b$ belong to $S_{3}$.
Therefore the required probability $=\frac{{ }^{10} C_{2}+{ }^{10} C_{1}{ }^{10} C_{1}}{{ }^{10} C_{3}}=\frac{45+100}{(15)(29)}=\frac{1}{3}$.