The range of a random variable $X$ is ${0, 1, 2}$ . Given that $P (X = 0) = 3 C^{3}, P (X = 1) = 4 C - 10 C^{2}, P (X = 2) = 5 C - 1$
i) Find the value of $C$ .
ii) $P (X < 1), P (1 < X \leq 2)$ and $P (0 < X \leq 3)$

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Solution

(i) We must have $\sum P (X) = 1$
$\Rightarrow P (X = 0) + P (X = 1) + P (X = 2) = 1$
$\Rightarrow (3 C^{3}) + (4 C - 10 C^{2}) + (5 C - 1) = 1$
$\Rightarrow 3 C^{3} - 10 C^{2} + 9 C - 2 = 0$
$\Rightarrow (C - 1) (3 C^{2} - 7 C + 2) = 0$
$\Rightarrow (C - 1) (C - 2) (3 C - 1) = 0$
$\Rightarrow C = 1, 2, \frac{1}{3}$
Also $0 \leq P (X = 0), P (X = 1), P (X = 2) \leq 1$
So only acceptable value of $C$ is $\frac{1}{3}$ .
(ii) $P (X < 1) = P (X = 0) = 3 {(\frac{1}{3})}^{3} = \frac{1}{9}$
(iii) $P (1 < X \leq 2) = P (X = 2) = 4 C - 10 C^{2} = 4 \cdot \frac{1}{3} - 10 \cdot \frac{1}{9} = \frac{2}{9}$
And (iv) $P (0 < X \leq 3) = 1 - P (X = 0) = 1 - \frac{1}{9} = \frac{8}{9}$

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