2002 AMC 12B Problem 16

Question

Juan rolls a fair regular octahedral die marked with the numbers

1

through

8.

Then Amal rolls a fair six-sided die. What is the probability that the product of the two rolls is a multiple of

3?

\dfrac{1}{12}

\dfrac{1}{3}

\dfrac{1}{2}

\dfrac{7}{12}

\dfrac{2}{3}

The product is a multiple of

3

exactly when at least one roll is

3

or

6

Compute

1

minus the probability that neither die shows a

3

or

6

The product is a multiple of

3

if and only if at least one die shows

3

or

6.

The octahedral die avoids

3,6

with probability

\dfrac68=\dfrac34,

and the six-sided die avoids them with probability

\dfrac46=\dfrac23.

So neither shows a multiple of

3

with probability

\dfrac34\cdot\dfrac23=\dfrac12,

and the answer is

1-\dfrac12=\dfrac12.

Thus, the correct answer is C.

Problem 16 in other years