Answer
a. $P(0 \leq X \leq 2)=\frac{1}{3}$
b. $P(X\gt 2)=\frac{2}{3}$
c. $P(1 \leq X \leq 3)=\frac{295}{432}$
Work Step by Step
We are given $\frac{x^{3}}{12}$ if $0 \leq x \leq2$
and $f(x)=\frac{16}{3x^{3}}$ if $x \gt 2$
$\int^{2}_{0}\frac{1}{12}x^{3}dx + \int^{\infty}_{2} \frac{16}{3}\frac{1}{x^{3}}dx$
$=\frac{1}{12}\frac{x^{4}}{4}|^{2}_{0}+\frac{16}{3}\frac{x^{-2}}{-2}|^{\infty}_{2})$
$=\frac{1}{12}4+\frac{16}{3}\frac{1}{8}$
$=\frac{1}{3}+\frac{2}{3}=1$
a. $P(0 \leq X \leq 2)=\frac{1}{3}$
b. $P(X\gt 2)=\frac{2}{3}$
c. $P(1 \leq X \leq 3)=\int^{2}_{1}\frac{1}{12}x^{3}dx + \int^{3}_{2} \frac{16}{3}\frac{1}{x^{3}}dx$
$=\frac{1}{12}\frac{x^{4}}{4}|^{2}_{1}+\frac{16}{3}\frac{x^{-2}}{-2}|^{3}_{2})$
$=\frac{1}{12}\frac{15}{4}+\frac{16}{3}\frac{5}{72}$
$=\frac{5}{16}+\frac{10}{27}=\frac{295}{432}$
