2004 AMC 8 Problems/Problem 21

Problem

Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?

[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1));  draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5));   label("$A$",(-1,1)); label("$B$",(2,1));  label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5));  [/asy]

$\textbf{(A)}\ \frac14\qquad \textbf{(B)}\ \frac13\qquad \textbf{(C)}\ \frac12\qquad \textbf{(D)}\ \frac23\qquad \textbf{(E)}\ \frac34$

Solution

An even number comes from multiplying an even and even, even and odd, or odd and even. Since an odd number only comes from multiplying an odd and odd, there are less cases and it would be easier to find the probability of spinning two odd numbers from $1$. Multiply the independent probabilities of each spinner getting an odd number together and subtract it from $1$.

\[1-\frac24 \cdot \frac23 = 1- \frac13 = \boxed{\textbf{(D)}\ \frac23}\]


Solution 2

We can make a chart and the we see that the 12 possibilities: 1, 2, 3, 2, 4, 6, 3, 6, 9, 4, 8, and 12. Out of these only 8 work; thus the probability is \[\boxed{\textbf{(D)}\ \frac23}\]

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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