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

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

First, we find k using $$\sum_{x=0}^{\infty} P(X = x) = 1$$.

$$\sum_{x=0}^{\infty} k(x+1)3^{-x} = k \sum_{x=0}^{\infty}(x+1)\left(\frac{1}{3}\right)^x = k \sum_{n=1}^{\infty} n \left(\frac{1}{3}\right)^{n-1}$$

Using the formula $$\sum_{n=1}^{\infty} n r^{n-1} = \frac{1}{(1-r)^2}$$ for $$|r| < 1$$:

$$= k \cdot \frac{1}{(1-1/3)^2} = k \cdot \frac{1}{(2/3)^2} = k \cdot \frac{9}{4}$$

Setting this equal to 1: $$k = \frac{4}{9}$$

Next, we find $$P(X \geq 2) = 1 - P(X=0) - P(X=1)$$.

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

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

$$P(X \geq 2) = 1 - \frac{4}{9} - \frac{8}{27} = 1 - \frac{12}{27} - \frac{8}{27} = 1 - \frac{20}{27} = \frac{7}{27}$$

The correct answer is Option 1: $$\frac{7}{27}$$.