Question

The random variable $X$ has a probability distribution $P\left(X\right)$ of the following form,
$P\left(X\right)=\{\begin{array}{cc}k,& x=0\\ 2k,& x=1\\ 3k,& x=2\\ 0,& \text{otherwise}\end{array}$
then value of $k$ is:

• A. $\frac{1}{2}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{6}$

Given: A random variable $X$ with its probability distribution.
As we know the sum of all the probabilities in a probability variable must be one.
i.e.$\sum _{i=1}^n{p}_i=1,\text{where}{p}_i>0\text{and}i=0,1,2,...,n$
Hence the sum of probabilities of given table:
$\Rightarrow k+2k+3k+0=1$
$\Rightarrow 6k=1\Rightarrow k=\frac{1}{6}$