If the probability distribution of a random variable X is x 1 2 3 4 5  P( X = x)    k    2k   3k   3k    k   Value of P(X > 2) is

$\frac{5}{6}$ $\frac{9}{10}$ $\frac{4}{5}$ $\frac{7}{10}$

$\frac{7}{10}$

Here k + 2k + 3k + 3k + k = 1 $\Rightarrow k=\frac{1}{10}$ so  P(X > 2) = 3k + 3k + k = 7k = $\frac{7}{10}$