A random variable $X$ has the following probability distribution:
$X$
$0$
$1$
$2$
$3$
$4$
$5$
$6$
$7$
$P (X)$
$0$
$k$
$2 k$
$2 k$
$3 k$
$k^{2}$
$2 k^{2}$
$7 k^{2} + k$
Determine
(i) $k$
(ii) $P (X < 3)$
(iii) $P (X > 6)$
(iv) $P (0 < X < 3)$

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Solution

(i) Since X is random variable.
$\sum P (X_{i}) = 1$
$0 + k + 2 k + 2 k + 3 k + k^{2} + 2 k^{2} + 7 k^{2} + k = 1 ⟹ 10 k^{2} + 9 k - 1 = 0 ⟹ 10 k^{2} + 10 k - k - 1 = 0 ⟹ 10 k (k + 1) - 1 (k + 1) = 0 ⟹ (k + 1) (10 k - 1) = 0 ⟹ k = - 1, k = \frac{1}{10}$
$k = - 1$ not possible, hence $k = \frac{1}{10}$
(ii) $P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2) ⟹ = 0 + k + 2 k = 3 k ⟹ P (X < 3) = 3 \times \frac{1}{10} = \frac{3}{10}$
(iii) $P (X > 6) = P (X = 7) = 7 k^{2} + k = \frac{7}{100} + \frac{1}{10} ⟹ P (X > 6) = \frac{7 + 10}{100} = \frac{17}{100}$
(iv) $P (0 < X < 3) = P (X = 1) + P (X = 2) = k + 2 k = 3 k ⟹ P (0 < X < 3) = \frac{3}{10}$

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