Question

Two numbers $x$ and $y$ are chosen such that $x\in [0,4],y\in [0,4]$ then probability $y^2\le x$

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{1}{3}$

$\frac{{\int }_0^4{y}^{2}dy}{2}=\frac{32}{3}$

$Probability=\frac{\frac{32}{3}}{4\times 4}=\frac{2}{3}$