Problem 597

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.