The probability distribution of a random variable X is given belo… - Vidyadip

Question

The probability distribution of a random variable X is given below:

$\text{X}$	$0$	$1$	$2$	$3$
$\text{P}(\text{X})$	$\text{k}$	$\frac{\text{k}}{2}$	$\frac{\text{k}}{4}$	$\frac{\text{k}}{8}$

Determine the value of k.
Determine $\text{P}(\text{X}\leq2)$ and $\text{P}(\text{X}\geq2)$
Find $\text{P}(\text{X}\leq2)+\text{P}(\text{X}\geq2)$

✓

Answer

We have,

$\text{X}$	$0$	$1$	$2$	$3$
$\text{P}(\text{X})$	$\text{k}$	$\frac{\text{k}}{2}$	$\frac{\text{k}}{4}$	$\frac{\text{k}}{8}$

Since, $\sum_\limits{\text{i}=1}^\text{n}\text{P}_{\text{i}}=1,\text{i}=1,2, ....,\text{n}$ and $\text{P}_{\text{i}}\geq0$

$\therefore\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}+\frac{\text{k}}{8}=1$

$\Rightarrow8\text{k}+4\text{k}+2\text{k}+\text{k}=8$

$\therefore\text{k}=\frac{8}{15}$

$\text{P}(\text{X}\leq2)=\text{P}(0)+\text{P}(1)+\text{P}(2)$

$=\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}$

$=\frac{(4\text{k}+2\text{k}+\text{k})}{4}=\frac{7\text{k}}{4}$

$=\frac{7}{4}\cdot\frac{8}{15}=\frac{14}{15}$

And $\text{P}(\text{X}\geq2)=\text{P}(3)=\frac{\text{k}}{8}$

$=\frac{1}{8}\cdot\frac{8}{15}=\frac{1}{15}$

$\text{P}(\text{X}\leq2)+\text{P}(\text{X}\geq2)$

$=\frac{14}{15}+\frac{1}{15}=1$

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