The p. d. f. of a continuous random variable $\mathrm{X}$ is given by $f(x)=\frac{1}{2} \quad$ if $\quad 0…
The p. d. f. of a continuous random variable $\mathrm{X}$ is given by
$f(x)=\frac{1}{2} \quad$ if $\quad 0 < x < 2$
$=0 \quad$ otherwise
and if $a=P\left(X < \frac{1}{2}\right), b=P\left(X>\frac{3}{2}\right)$, then relation between $a$ and $b$ is
$a-b=0$
$2a-b=0$
$3a-b=0$
$a-2 b=0$
Solution
$\begin{aligned} a &=PX < \frac{1}{2}=\int f(x) d x \\ &=\int f(x) d x-\int f(x) d x=0+\int \frac{1}{2} d x \\ &=\frac{1}{2}[x] \frac{1}{2}=\frac{1}{2} \frac{1}{2}-0 \quad=\frac{1}{4}=a=\frac{1}{4} \\ b &=P\left(X>\frac{3 a}{2}=\int f(x) d x\right.\\ &=\int_{\frac{3}{2}}^{2} f(x) d x+\int_{2}^{\infty} f(x) d x=\int_{\frac{3}{2}}^{2} \frac{1}{2} d x+0 \\ &=\frac{1}{2}[x]_{\frac{3}{2}}^{2}=\frac{1}{2}\left(2-\frac{3}{2}\right)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \Rightarrow b=\frac{1}{4} \\ \therefore & a-b=\frac{1}{4}-\frac{1}{4}=0 \quad \end{aligned}$