1997 AHSME Problems/Problem 10

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Problem

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 an odd number?

$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{4}{9}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{5}{9}\qquad\textbf{(E)}\ \frac{11}{18}$

Solution

On the first die, the chance of an odd number is $\frac{4}{6} = \frac{2}{3}$, and the chance of an even number is $\frac{1}{3}$.

On the second die, the chance of an odd number is $\frac{1}{3}$, and the chance of an even number is $\frac{2}{3}$.

To get an odd sum, we need exactly one even and one odd.

The odds of the first die being even and the second die being odd is $\frac{1}{3}\cdot\frac{1}{3} = \frac{1}{9}$

The odds of the first die being odd and the second die being even is $\frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}$.

The overall probability is $\frac{1}{9} + \frac{4}{9} = \frac{5}{9}$, and the answer is $\boxed{D}$.

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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