2015 AMC 8 Problems/Problem 7

Revision as of 10:24, 16 December 2015 by Ghghghghghghghgh (talk | contribs) (Solution 2)

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?

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


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 $\left ( \frac{2}{3} \right )^2=\frac{4}{9}$ probability 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

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

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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