Find the value of P(Y >= X | Y >= 12)

Question

$X$ and $Y$ are independent continuous random variables each with uniform distribution from $[0,100]$. I know that $$ \mathbb{P}[Y \ge 12] = 1 - \mathbb{P}[Y \le 12] = \frac{88}{100} $$ I'm just confused on how to find $\mathbb{P}[Y \ge X \cap Y \ge 12]$.

Answer 1

Since this is a uniform distribution, $P(Y\geq X)=\frac{Y-0}{100}=\frac Y{100}$.

So, we have $$P(Y\geq X)\cap P(Y\geq12)=\frac1{100}\int_{12}^{100}\frac Y{100}dY=\frac1{20000}Y^2\bigg\vert_{12}^{100}=0.4928$$