Probability Distributions and Probability 
Suppose $X \sim N(3, 4)$, and let $Y = X^2$. Find $\Pr(Y ≥ 12)$.

What does $Y$ mean?
 A: $Y$ means just what you said it means: It's the square of $X$, i.e. $Y=X^2$.  Notice that the event that $Y\ge 12$ is the same as the event $\left[X\ge\sqrt{12}\text{ or }X\le-\sqrt{12}\right]$, so the probability that $Y\ge 12$ is the same as the probability of the latter event.
PS: Because of some confusion appearing from comments below, I am adding some more details here.  The probability is
$$
\Pr\left( \frac{X-3}{2} \ge \frac{\sqrt{12}-3}2 \right) + \Pr\left( \frac{X-3}{2} \le \frac{-\sqrt{12}-3}2 \right)
$$
$$
= 1-\Phi\left( \frac{\sqrt{12}-3}2 \right) + \Phi\left(\frac{-\sqrt{12}-3}{2}\right) 
$$
$$
\approx 1-0.59175 + 0.00061 = 0.40886
$$
A: $Y$ is a transformation of $X$, and $Y/\sigma^2$ is Noncentral Chisquare distributed, where $\lambda=\mu^2/\sigma^2=3^2/4=2.25$, $\,\,k=1$.
Its CDF $F^{\lambda,k}_{Y/\sigma^2}(x)$ is a bit complicated, you may use MATLAB for it.
Then:
$P(Y\geq12)=P(Y/\sigma^2\geq12/\sigma^2)=1-F^{\lambda,k}_{Y/\sigma^2}(12/\sigma^2)\approx 1-0.591136=0.408864$
