2015 AMC 8 Problems/Problem 7

Each of two boxes contains three chips numbered $1$ , $2$ , $3$ . A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?

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

Solution

We can instead find the probability that their product is odd, and subtract this from $1$ . In order to get an odd product, we have to draw an odd number from each box. We have a $\frac{2}{3}$ probability of drawing an odd number from one box, so there is a ${\frac{2}{3}}^2=\frac{4}{9}$ of having an odd product. Thus, there is a $1-\frac{4}{9}=\frac{5}{9}$ probability of having an even product. We get our answer to be $\boxed{\textbf{(E) }\frac{5}{9}}$ .

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2015 AMC 8 Problems/Problem 7

Solution

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