Probability — Class 12 Maths Solution

.: Since $\sum P (X) = 1$

$\therefore$ $0 + k + 2k + 2k + 3k + {k^2} + 2{k^2} + 7{k^2} + k = 1$

$\Rightarrow 10{k^2} + 9k - 1 = 0 \Rightarrow k = \cfrac{{ - 9 \pm \sqrt {81 + 40} }}{{20}} = \cfrac{{ - 9 \pm 11}}{{20}} = \cfrac{1}{{10}}, - 1$

Since the probability of the event lies between and 1, there fore rejecting

$k = - 1 \Rightarrow k = \cfrac{1}{{10}}$

(ii) $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

(from (i) )
(iii) $P(X > 6) = P(7)$

(from
(iv) $P(0 < X < 3) = P(X = 1) + P(X = 2)$
(from (i))