# 2006 AMC 8 Problems/Problem 17

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

Jeff rotates spinners $P$, $Q$ and $R$ and adds the resulting numbers. What is the probability that his sum is an odd number?

$[asy] size(200); path circle=circle((0,0),2); path r=(0,0)--(0,2); draw(circle,linewidth(1)); draw(shift(5,0)*circle,linewidth(1)); draw(shift(10,0)*circle,linewidth(1)); draw(r,linewidth(1)); draw(rotate(120)*r,linewidth(1)); draw(rotate(240)*r,linewidth(1)); draw(shift(5,0)*r,linewidth(1)); draw(shift(5,0)*rotate(90)*r,linewidth(1)); draw(shift(5,0)*rotate(180)*r,linewidth(1)); draw(shift(5,0)*rotate(270)*r,linewidth(1)); draw(shift(10,0)*r,linewidth(1)); draw(shift(10,0)*rotate(60)*r,linewidth(1)); draw(shift(10,0)*rotate(120)*r,linewidth(1)); draw(shift(10,0)*rotate(180)*r,linewidth(1)); draw(shift(10,0)*rotate(240)*r,linewidth(1)); draw(shift(10,0)*rotate(300)*r,linewidth(1)); label("P", (-2,2)); label("Q", shift(5,0)*(-2,2)); label("R", shift(10,0)*(-2,2)); label("1", (-1,sqrt(2)/2)); label("2", (1,sqrt(2)/2)); label("3", (0,-1)); label("2", shift(5,0)*(-sqrt(2)/2,sqrt(2)/2)); label("4", shift(5,0)*(sqrt(2)/2,sqrt(2)/2)); label("6", shift(5,0)*(sqrt(2)/2,-sqrt(2)/2)); label("8", shift(5,0)*(-sqrt(2)/2,-sqrt(2)/2)); label("1", shift(10,0)*(-0.5,1)); label("3", shift(10,0)*(0.5,1)); label("5", shift(10,0)*(1,0)); label("7", shift(10,0)*(0.5,-1)); label("9", shift(10,0)*(-0.5,-1)); label("11", shift(10,0)*(-1,0)); [/asy]$

$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$

## Solution

In order for Jeff to have an odd number sum, the numbers must either be Odd + Odd + Odd or Even + Even + Odd. We easily notice that we cannot obtain Odd + Odd + Odd because spinner $Q$ contains only even numbers. Therefore we must work with Even + Even + Odd and spinner $Q$ will give us one of our even numbers. We also see that spinner $R$ only contains odd, so spinner $R$ must give us our odd number. We still need one even number from spinner $P$. There is only 1 even number: $2$. Since spinning the required numbers are automatic on the other spinners, we only have to find the probability of spinning a $2$ in spinner $P$, which is clearly $\boxed{\textbf{(B)}\ \frac{1}{3}}$