Question

The random variable X has a probability distribution P(X) of the following form, where k is some number
$\{\begin{array}{c}K,ifX=0\\ 2k,ifX=1\\ 3k,ifX=2\\ 0,ifotherwise\end{array}$

Determine the value of k

$\{\begin{array}{c}K,ifX=0\\ 2k,ifX=1\\ 3k,ifX=2\\ 0,ifotherwise\end{array}$

Answer 1

Given distribution of X is

$\begin{array}{ccccc}X& 0& 1& 2& otherwise\\ P\left(X\right)& k& 2k& 3k& 0\end{array}$

Since, $\sum P\left(X\right)=1$, therefore P(0)+P(1)+P(2)+P(otherwise)=1

$\Rightarrow$ $k+2k+3k+0=1\Rightarrow 6k=1\Rightarrow k=\frac{1}{6}$

Given distribution of X is

$\begin{array}{ccccc}X& 0& 1& 2& otherwise\\ P\left(X\right)& k& 2k& 3k& 0\end{array}$​​​

$P(X<2)=P\left(0\right)+P\left(1\right)=k+2k=3k=3\times \frac{1}{6}=\frac{1}{2}$

$P(X\le 2)=P\left(0\right)+P\left(1\right)+P\left(2\right)=k+2k+3k=6k=6\times \frac{1}{6}=1$

and $P(X\ge 2)=P\left(2\right)+P\left(otherwise\right)=3k+0=3\times \frac{1}{6}=\frac{1}{2}$