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

We are given $$P(X = x) = k(x+1)3^{-x}$$ for $$x = 0, 1, 2, 3, \ldots$$

Since the total probability must equal 1: $$\displaystyle\sum_{x=0}^{\infty} k(x+1)3^{-x} = 1$$.

We know that $$\displaystyle\sum_{x=0}^{\infty} (x+1)r^x = \dfrac{1}{(1-r)^2}$$ for $$|r| \lt 1$$.

With $$r = 1/3$$: $$\displaystyle\sum_{x=0}^{\infty} (x+1)\left(\dfrac{1}{3}\right)^x = \dfrac{1}{(1 - 1/3)^2} = \dfrac{1}{(2/3)^2} = \dfrac{9}{4}$$.

So $$k \cdot \dfrac{9}{4} = 1$$, giving $$k = \dfrac{4}{9}$$.

Now, $$P(X \ge 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)$$.

$$P(X = 0) = \dfrac{4}{9} \cdot 1 \cdot 1 = \dfrac{4}{9}$$

$$P(X = 1) = \dfrac{4}{9} \cdot 2 \cdot \dfrac{1}{3} = \dfrac{8}{27}$$

$$P(X = 2) = \dfrac{4}{9} \cdot 3 \cdot \dfrac{1}{9} = \dfrac{12}{81} = \dfrac{4}{27}$$

$$P(X = 0) + P(X = 1) + P(X = 2) = \dfrac{4}{9} + \dfrac{8}{27} + \dfrac{4}{27} = \dfrac{12}{27} + \dfrac{8}{27} + \dfrac{4}{27} = \dfrac{24}{27} = \dfrac{8}{9}$$

Therefore, $$P(X \ge 3) = 1 - \dfrac{8}{9} = \dfrac{1}{9}$$.

Hence, the correct answer is Option D.