Question

Two real numbers $x,y$ are chosen from the interval $[0,8]$ then the probability that $y^2>2x$ is

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{2}{9}$

Answer 1

The correct option is C $\frac{2}{3}$

Total Area $=8\times 8=64$
Favourable Area $=64-{\int }_0^8\sqrt{2}\sqrt{x}dx=64-\frac{2\sqrt{2}}{3}[8^{\frac{3}{2}}-0]=64-\frac{64}{3}=\frac{2}{3}\times 64$
Required probability $=\frac{64\times \frac{2}{3}}{64}=\frac{2}{3}$