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

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