Difference between revisions of "2002 AMC 12B Problems/Problem 16"

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[[Category:Introductory Algebra Problems]]
 
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Revision as of 10:22, 4 July 2013

Problem

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?

$\mathrm{(A)}\ \frac 1{12} \qquad\mathrm{(B)}\ \frac 13 \qquad\mathrm{(C)}\ \frac 12 \qquad\mathrm{(D)}\ \frac 7{12} \qquad\mathrm{(E)}\ \frac 23$

Solution

Solution 1

On both dice, only the faces with the numbers $3,6$ are divisible by $3$. Let $P(a) = \frac{2}{8} = \frac{1}{4}$ be the probability that Juan rolls a $3$ or a $6$, and $P(b) = \frac{2}{6} = \frac 13$ that Amal does. By the Principle of Inclusion-Exclusion,

\[P(a \cup b) = P(a) + P(b) - P(a \cap b) = \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{2} \Rightarrow \mathrm{(C)}\]

Alternatively, the probability that Juan rolls a multiple of $3$ is $\frac{1}{4}$, and the probability that Juan does not roll a multiple of $3$ but Amal does is $\left(1 - \frac{1}{4}\right) \cdot \frac{1}{3} = \frac{1}{4}$. Thus the total probability is $\frac 14 + \frac 14 = \frac 12$.

Solution 2

The probability that neither Juan nor Amal rolls a multiple of $3$ is $\frac{6}{8} \cdot \frac{4}{6} = \frac{1}{2}$; using complementary counting, the probability that at least one does is $1 - \frac 12 = \frac 12 \Rightarrow \mathrm{(C)}$.

See also

2002 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 12 Problems and Solutions

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