Question
A random variable X has the following probability distribution:
Values of X: 0 1 2 3 4 5 6 7 8
P(X) a 3a 5a 7a 9a 11a 13a 15a 17a
Determine:
$\text{P}(\text{X}<3),\text{P}(\text{X}\geq3),\text{P}(0<\text{X}<5).$

Answer

$\text{P}(\text{X}<3)=\text{P}(0)+\text{P}(1)+\text{P}(2)$
$=\text{a}+3\text{a}+5\text{a}$
$=9\text{a}$
$=9\Big(\frac{1}{81}\Big)$
$\therefore\ \text{P}(\text{X}<3)=\frac{1}{9}$
$\text{P}(\text{X}\geq3)=1-\text{P}(\text{X}<3)=1-\frac{1}{9}=\frac{8}{9}$
$\text{P}(0<\text{X}<5)=\text{P}(1)+\text{P}(2)+\text{P}(3)+\text{P}(4)$
$=3\text{a}+5\text{a}+7\text{a}+9\text{a} $
$=24\text{a}$
$=24\Big(\frac{1}{81}\Big)$
$\therefore\ \text{P}(0<\text{X}<5)=\frac{8}{27}$

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