Question

The pdf of a random variable $X$ is $f\left(x\right)=3\left(1-2x^{2}\right)$, $0 < x < 1$
$=0\quad$ otherwise
The $P\left(\frac{1}{4} < X < \frac{1}{3}\right)=$ ...

• A. $\frac{179}{864}$
• B. $\frac{159}{864}$
• C. $\frac{169}{864}$
• D. $\frac{189}{864}$

Answer 1

Answer: (A)

We have, p.d.f of a random variable
$X$ is $f(x)=3\left(1-2 x^{2}\right),0 < x < 1$
$=0$, otherwise
$\therefore P\left(\frac{1}{4} < X < \frac{1}{3}\right)=\int_\limits{1 / 4}^{1 / 3} f(x) d x$
$=\int_\limits{1 / 4}^{1 / 3} 3\left(1-2 x^{2}\right) d x$
$=3\left[x-\frac{2}{3} x^{3}\right]_{1 / 4}^{1 / 3}$
$=3\left[\left(\frac{1}{3}-\frac{2}{3}\left(\frac{1}{3}\right)^{3}\right)-\left(\frac{1}{4}-\frac{2}{3}\left(\frac{1}{4}\right)^{3}\right)\right]$
$=3\left[\left(\frac{1}{3}-\frac{1}{4}\right)-\frac{2}{3}\left(\frac{1}{3^{3}}-\frac{1}{4^{3}}\right)\right]$
$=3\left[\frac{1}{12}-\frac{2}{3} \times \frac{37}{1728}\right]$
$=3 \times \frac{179}{2592}$
$=\frac{179}{864}$