Question

The random variable $X$ has a probability distribution $P(X)$ of the following form, where $k$ is some number.
$P(X)=\begin{cases} k, \text { if } x=0 \\ 2 k, \text { if } x=1 \\ 3 k, \text { if } x=2 \\ 0, \text { otherwise } \end{cases}$
Then, match the terms of column I with their respective values in column II and choose the correct option from the codes given below.

Column I		Column II
A	The value of $k$	1	$\frac{1}{2}$
B	$P(X< 2)$	2	1
C	$P(X \leq 2)$	3	$\frac{1}{2}$
D	$P(X \geq 2)$	4	$\frac{1}{6}$

• A. A- 4 ,B- 2 ,C- 3 ,D- 1
• B. A-4 ,B- 3 ,C-2 ,D- 1
• C. A-3 ,B- 1 ,C-4 ,D- 2
• D. A- 4,B- 2 ,C-1 ,D- 3

Answer 1

Answer: (B)

Given distribution of $X$ is

A. Since, $\Sigma P(X)=1$,
$P(0)+P(1)+P(2)+P \text { (otherwise) } =1 $
$\rightarrow k+2 k+3 k+0 =1 $
$\rightarrow 6 k =1 $
$\rightarrow k =\frac{1}{6}$
B. $P(X<2)=P(0)+P(1)$
$= k+2 k=3 k=3 \times \frac{1}{6}=\frac{1}{2}$
C. $P(X \leq 2)=P(0)+P(1)+P(2)$
$= k+2 k+3 k=6 k=6 \times \frac{1}{6}=1$
D. $P(X \geq 2)=P(2)+P$ (otherwise) $=3 k+0=3 \times \frac{1}{6}=\frac{1}{2}$