Difference between revisions of "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 $\boxed{\textbf{(E) }\frac{5}{9}}$.

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