Question

A random variable $X$ has the following probability distribution:
$\begin{array}{ccccccccc}\text{X}& \text{0}& \text{1}& \text{2}& \text{3}& \text{4}& \text{5}& \text{6}& \text{7}\\ \text{P(X)}& \text{0}& \text{k}& \text{2k}& \text{2k}& \text{3k}& k^2& 2k^2& 7k^2+k\end{array}$
Then value of $k$ is:

• A. $\frac{1}{10}$
• B. $\frac{1}{5}$
• C. $-1$
• D. $\frac{1}{7}$

Answer 1

The correct option is A $\frac{1}{10}$

Given a random variabe $X$ with its probability distribution.
As we know the sum of all the probabilities in a probability distribution of a random variable must be one.
i.e $\sum _{i=1}^n{p}_i=1,\text{where}{p}_i>0$ and $i=0,1,2,...,n$
Hence the sum of probabilities of given table :
$=0+k+2k+2k+3k+k^2+2k^2+7k^2+k=1$
$\Rightarrow 10k^2+9k=1$
$\Rightarrow 10k^2+9k-1=0$
$\Rightarrow (10k-1)(k+1)=0$
$\Rightarrow k=-1,\frac{1}{10}$
It is known that probability of any observation must always be positive and it can't be negative.
So $k=\frac{1}{10}$