If the probability that the random variable X takes values x is...

Given,

The probability that the random variable $X$ takes values $x$ is given by $P ( X = x ) = k ( x + 1 ) {3}^{- x}$, $x = 0, 1, 2, 3, \ldots \ldots$, where $k$ is a constant,

Now we know that,

$P ( X = 0 ) + P ( x = 1 ) + P ( x = 2 ) + \ldots \ldots \ldots = 1$

$⇒ \dfrac{k}{{3}^{0}} + \dfrac{2k}{{3}^{1}} + \dfrac{3k}{{3}^{2}} + \ldots . . = 1$

$⇒ k\left(1 + \dfrac{2}{3} + \dfrac{3}{{3}^{2}} + \ldots .\right) = 1$

Now finding,

$S = 1 + \dfrac{2}{3} + \dfrac{3}{{3}^{2}} + \dfrac{4}{{3}^{3}} + \ldots . . . . \left(1\right)$

$⇒ \dfrac{S}{3} = \dfrac{1}{3} + \dfrac{2}{{3}^{2}} + \dfrac{3}{{3}^{2}} + \ldots \ldots . . . . . \left(2\right)$

Now on subtracting above two equation we get,

$⇒ \dfrac{2S}{3} = 1 + \dfrac{1}{3} + \dfrac{1}{{3}^{2}} + \ldots \ldots$

$⇒ S = \dfrac{9}{4}$

Hence, $k\left(1 + \dfrac{2}{3} + \dfrac{3}{{3}^{2}} + \ldots .\right) = 1$

$⇒ k = \dfrac{4}{9}$

Now finding,

$P ( X \ge 2 ) = P ( 2 ) + P ( 3 ) + \ldots \ldots$

$⇒ P ( X \ge 2 ) = 1 - P ( 0 ) - P ( 1 )$

$⇒ P ( X \ge 2 ) = 1 - \left(\dfrac{k}{1} + \dfrac{2k}{3}\right) = 1 - \dfrac{20}{27} = \dfrac{7}{27}$