2015 AMC 8 Problems/Problem 7

Revision as of 22:22, 13 October 2020 by Vibha ganji (talk | contribs) (Solution 3)

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 1

We can instead calculate 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

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

Solution 3

We can also list out the numbers. Hat A has chips $1$, $2$, and $3$, and Hat B also has chips $1$, $2$, and $3$. Chip $1$(from Hat A)

could be with 3 partners from Hat B. This is also the same for chips $2$ and $3$ from Hat A. $3+3+3=9$ total sums. Chip $1$ could be multiplied with 2 other chips to make an even product, just like chip $3$. Chip $2$ can only multiply with 1 chip. $2+2+1=5$. The answer is $\boxed{\textbf{(E) }\frac{5}{9}}$.

Solution 4

Here is another way:

Let's start by finding the denominator: Total choices. There are $3$ chips we can choose from in the 1st box, and $3$ chips we can choose from in the 2nd box. We do $3*3$, and get $9$. Now - to find the numerator: Desired choices. To get an even number, we need to pick 2 from at least one of the boxes. There are $2$ choices as to finding which box we will draw the 2 from. Then we have $3$ choices from the other box to pick any of the other chips, $1, 2, 3$.

$\frac{3 \cdot2}{9} = \frac{6}{9}$ However, we are over counting, the $(2,2)$ configuration twice, and so we subtract that one configuration from our total. $\frac{6}{9} - \frac{1}{9}$. Thus, our answer is $\boxed{\textbf{(E) }\frac{5}{9}}$.

Solution 4 by: del-math.

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

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

Invalid username
Login to AoPS