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GO Classes DPP | GATE DA | Probability | Continuous Random Variables

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2 2 votes

Let $X$ be a continuous random variable with probability density function

$$
f(x)= \begin{cases}cx^2& , x \in[0,1] \\\\ c(2-x) & , x \in(1,2]\\\\0 & , \text { otherwise }\end{cases}
$$

where $c\in\mathbb{R}$. What is $P\left(X\le 1 \mid X\le \frac{3}{2}\right)$?

  1. $\frac{6}{17}$
     
  2. $\frac{8}{17}$
     
  3. $\frac{10}{17}$
     
  4. $\frac{12}{17}$

1 Answer

0 0 votes

Since $f(x)$ is a probability density function,

$\int_0^1 cx^2dx+\int_1^2 c(2-x)dx=1$

$\Rightarrow c\left(\int_0^1 x^2dx+\int_1^2 (2-x)dx\right)=1$

$\Rightarrow c\left(\frac{1}{3}+\frac{1}{2}\right)=1$

$\Rightarrow c\cdot\frac{5}{6}=1$

$\Rightarrow c=\frac{6}{5}$


Now,

$P(X\le 1)=\int_0^1 \frac{6}{5}x^2,dx=\frac{6}{5}\cdot\frac{1}{3}=\frac{2}{5}$

Also,

$P\left(X\le \frac{3}{2}\right)=\int_0^1 \frac{6}{5}x^2,dx+\int_1^{3/2}\frac{6}{5}(2-x),dx$

$=\frac{2}{5}+\frac{6}{5}\int_1^{3/2}(2-x),dx$


Now,

$\int_1^{3/2}(2-x),dx=\left[2x-\frac{x^2}{2}\right]_1^{3/2}$

$=\left(3-\frac{9}{8}\right)-\left(2-\frac{1}{2}\right)$

$=\frac{15}{8}-\frac{12}{8}=\frac{3}{8}$


So,

$P\left(X\le \frac{3}{2}\right)=\frac{2}{5}+\frac{6}{5}\cdot\frac{3}{8}$

$=\frac{2}{5}+\frac{9}{20}=\frac{8}{20}+\frac{9}{20}=\frac{17}{20}$


Therefore,

$P\left(X\le 1 \mid X\le \frac{3}{2}\right)=\frac{P(X\le 1)}{P\left(X\le \frac{3}{2}\right)}$

$=\frac{2/5}{17/20}=\frac{2}{5}\cdot\frac{20}{17}=\frac{8}{17}$


So the correct answer is B.

$\boxed{P\left(X\le 1 \mid X\le \frac{3}{2}\right)=\frac{8}{17}}$

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