A random variable $X$ has the following probability distribution:
Determine
(i) $k$
(ii) $P (X < 3)$
(iii) $P (X > 6)$
(iv) $P (0 < X < 3)$

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Solution

(i) It is known that the sum of probabilities of a probability distribution of random variables is one.
$∴ 0 + k + 2 k + 3 k + k^{2} + 2 k^{2} + (7 k^{2} + k) = 1$
$\Rightarrow 10 k^{2} + 9 k - 1 = 0$
$\Rightarrow (10 k - 1) (k + 1) = 0$
$\Rightarrow k = - 1, \frac{1}{10}$
$k = - 1$ is not possible as the probability of an event is never negative.
$∴ k = \frac{1}{10}$
(ii) $P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)$
$= 0 + k + 2 k$
$= 3 k$
$= 3 \times \frac{1}{10}$
$= \frac{3}{10}$
(iii) $P (X > 6) = P (X = 7)$
$= 7 k^{2} + k$
$= 7 \times {\frac{1}{10}}^{2} + \frac{1}{10}$
$= \frac{7}{100} + \frac{1}{10}$
$= \frac{17}{100}$
(iv) $P (0 < X < 3) = P (X = 1) + P (X = 2)$
$= k + 2 k$
$= 3 k$
$= 3 \times \frac{1}{10}$
$= \frac{3}{10}$

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