If $f(x)=\left\{\begin{array}{ccc}3\left(1-2 x^2\right) & ; & 0<x<1 \\ 0 & ; & \text { otherwise }\end{array}\right.$
is a probability density function of $X$, then $P\left(\frac{1}{4}<x<\frac{1}{3}\right)$ is
(c) : $P(a<x<b)=\int_a^b f(x) d x$
$
\therefore P\left(\frac{1}{4}<x<\frac{1}{3}\right)=\int_{1 / 4}^{1 / 3} 3\left(1-2 x^2\right) d x=\left(3 x-\frac{6 x^3}{3}\right)_{1 / 4}^{1 / 3}
$
$\begin{aligned} & \Rightarrow\left(3 x-2 x^3\right)_{1 / 4}^{1 / 3}=\frac{3 \times 1}{3}-\frac{2 \times 1}{27}-\frac{3}{4}+\frac{2}{64} \\ & =1-\frac{2}{27}-\frac{3}{4}+\frac{1}{32}=\frac{864-64-648+27}{864}=\frac{179}{864}\end{aligned}$
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