Probability — Class 12 Maths Solution

We have

(i) Since, $\sum\limits_{i = 1}^n {{P_i}} = 1,i = 1,2, \ldots ,n$

and ${P_i} \ge 0$

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

$\Rightarrow$ $8k + 4k + 2k + k = 8$

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

(ii) $P(X \le 2) = P(0) + P(1) + P(2) = k + \frac{k}{2} + \frac{k}{4}$

$= \frac{{(4k + 2k + k)}}{4} = \frac{{7k}}{4} = \frac{7}{4} \cdot \frac{8}{{15}} = \frac{{14}}{{15}}$

and $P(X > 2) = P(3) = \frac{k}{8} = \frac{1}{8} \cdot \frac{8}{{15}} = \frac{1}{{15}}$

(iii) $P(X \le 2) + P(X > 2) = \frac{{14}}{{15}} + \frac{1}{{15}} = 1$