Question

Given that $xϵ[0,1]$ and $yϵ[0,1]$. Let $A$ be the event of $(x,y)$ satisfying $y^2\le x$ and $B$ be the event of $(x,y)$ satisfying $x^2\le y$. Then

• A. $P(A\cap B)=\frac{1}{3}$
• B. $A,B$ are exhaustive
• C. $A,B$ are mutually exclusive
• D. $A,B$ are independent

Answer 1

Answer: (A)

$P(A\cap B)=\frac{1}{3}$

The required probability is given by the shaded area in the diagram since the total area is $1$.
Thus, the probability = ${\int }_0^1(\sqrt{x}-x^2)dx={|\frac{2}{3}{x}^{\frac{3}{2}}-\frac{x^3}{3}|}_0^1=\frac{1}{3}$
From the diagram, it is clear that $A$ and $B$ are neither exhaustive nor mutually exclusive.
Also, $P\left(A\right)={\int }_0^1\left(\sqrt{x}\right)dx={\left|\frac{2}{3}{x}^{\frac{3}{2}}\right|}_0^1=\frac{2}{3}$ and $P\left(B\right)={\int }_0^1\left(x^2\right)dx={\left|\frac{x^3}{3}\right|}_0^1=\frac{1}{3}$.
Thus, $P(A\bigcap B)\ne P\left(A\right)\ast P\left(B\right)$.
Hence, (a) is correct.