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 $1-\frac{4}{9}=\boxed{\textbf{(E)}~\frac{5}{9}}$ probability of having an even product.

Solution 2

You can also make this problem into a spinner problem. You have the first spinner with $3$ equally divided

sections, $1, 2$ and $3.$ You make a second spinner that is identical to the first, with $3$ equal sections of

$1$ , $2$ , and $3$ . If the first spinner lands on $1$ , to be even, it must land on two. You write down the first

combination of numbers $(1,2)$ . Next, if the spinner lands on $2$ , it can land on any number on the second

spinner. We now have the combinations of $(1,2) (2,1) (2,2) (2,3)$ . Finally, if the first spinner ends on $3$ , we

have $(3,2).$ Since there are $3*3=9$ possible combinations, and we have $5$ evens, the final answer is

$\boxed{\textbf{(E) }\frac{5}{9}}$ .

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2015 AMC 8 Problems/Problem 7

Solution

Solution 2

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