An integer $m$ is chosen at random from the list $-9,-7,-5,-3,-1,1,3,5,7,9$. The probability that $m^{4}>100$ is
(A) $\frac{1}{5}$
(B) $\frac{3}{10}$
(C) $\frac{1}{2}$
(D) $\frac{2}{5}$
(E) $\frac{3}{5}$
There are 10 numbers in the list.

We note that

$$
\begin{gathered}
(-1)^{4}=1^{4}=1 \quad(-3)^{4}=3^{4}=81 \quad(-5)^{4}=5^{4}=625 \\
(-7)^{4}=7^{4}=2401 \quad(-9)^{4}=9^{4}=6561
\end{gathered}
$$

Thus, if $m=-3,-1,1,3$, then $m^{4}100$.

In other words, there are exactly six numbers in the list whose fourth power is larger than 100. Thus, if $m$ is chosen at random from the list, the probability that $m^{4}>100$ is $\frac{6}{10}$ or $\frac{3}{5}$.

ANsWer: (E)