A special six-sided die is rolled. The probability of rolling a number that is a multiple of three is $\frac{1}{2}$. The probability of rolling an even number is $\frac{1}{3}$. A possibility for the numbers on the die is
(A) $1,2,3,5,5,6$
(D) $1,2,3,3,4,6$
(B) $1,2,3,3,5,6$
(E) $2,3,3,3,5,6$
(C) $1,2,3,4,6,6$
Using the special six-sided die, the probability of rolling a number that is a multiple of three is $\frac{1}{2}$.

Since $\frac{1}{2}$ of 6 is 3 , then exactly 3 numbers on the die must be multiples of 3 .

Since the probability of rolling an even number is $\frac{1}{3}$ and $\frac{1}{3}$ of 6 is 2 , then exactly 2 numbers on the die must be even.

The die in (A) has only 2 numbers that are multiples of 3 (3 and 6 ), and thus may be eliminated. The die in (C) has 4 numbers that are even $(2,4,6,6)$, and thus may be eliminated.

The die in (D) has 3 numbers that are even $(2,4,6)$, and thus may be eliminated.

The die in (E) has 4 numbers that are multiples of $3(3,3,3,6)$, and thus may be eliminated.

The die in (B) has exactly 3 numbers that are multiples of $3(3,3,6)$, and exactly 2 even numbers ( 2 and 6 ), and is therefore the correct answer.

ANSWER: (B)