1997 AMC 12 Problem 10

Question

Two six-sided dice are fair in the sense that each face is equally likely to turn up. However, one of the dice has the

4

replaced by

3

and the other die has the

3

replaced by

4.

When these dice are rolled, what is the probability that the sum is odd?

\frac13

\frac49

\frac12

\frac59

\frac{11}{18}

An odd sum requires one odd result and one even result

Count odd-labeled faces on each modified die, including repeated labels

The first modified die has

4

odd faces and

2

even faces; the second has

2

odd and

4

even. Thus the probability is

(\frac{4}{6})(\frac{4}{6})+(\frac{2}{6})(\frac{2}{6})

=\frac{20}{36}=\frac{5}{9}.

The correct answer is D.

Problem 10 in other years