1. Key Concepts and Formulas

• Probability Mass Function (PMF) Properties: For a discrete random variable $X$ taking values $x_0, x_1, x_2, \ldots$, its PMF $P(X=x)$ must satisfy two conditions:
  1. $P(X=x) \geq 0$ for all $x$.
  2. $\sum_{x} P(X=x) = 1$ (the sum of all probabilities must equal 1).
• Sum of an Infinite Arithmetic-Geometric Series: A series of the form $\sum_{x=0}^{\infty} (ax+b)r^x$ can often be summed. A specific case relevant here is $\sum_{x=0}^{\infty} (x+1)r^x = 1 + 2r + 3r^2 + \ldots = \frac{1}{(1-r)^2}$, provided $|r|<1$.
• Complement Rule of Probability: For any event $A$, the probability of $A$ occurring is $P(A) = 1 - P(A^c)$, where $A^c$ is the complement of $A$. This is particularly useful for calculating probabilities over an infinite range by converting them into finite sums.

2. Step-by-Step Solution

Step 1: Understand the Problem and Formulate a Strategy We are given the probability mass function (PMF) for a discrete random variable $X$: $P(X=x)=k(x+1) 3^{-x}$ for $x=0,1,2,3 \ldots$. Our goal is to find $P(X \geq 3)$. The first step in any such problem is to determine the unknown constant $k$. We will use the fundamental property that the sum of all probabilities for all possible values of $X$ must be equal to 1: $\sum_{x=0}^{\infty} P(X=x) = 1$. Once $k$ is found, we need to calculate $P(X \geq 3)$. Directly summing $P(X=3) + P(X=4) + \ldots$ would involve an infinite sum. To simplify this, we will use the complement rule: $P(X \geq 3) = 1 - P(X < 3)$. Since $X$ takes integer values, $X < 3$ means $X$ can take values $0, 1,$ or $2$. Thus, $P(X < 3) = P(X=0) + P(X=1) + P(X=2)$, which is a finite sum.

Step 2: Determine the Constant $k$ We apply the total probability rule: $\sum_{x=0}^{\infty} P(X=x) = 1$ Substitute the given PMF: $\sum_{x=0}^{\infty} k(x+1) 3^{-x} = 1$ Factor out the constant $k$: $k \sum_{x=0}^{\infty} (x+1) \left(\frac{1}{3}\right)^x = 1$ Let's evaluate the infinite series $S = \sum_{x=0}^{\infty} (x+1) \left(\frac{1}{3}\right)^x$. Let $r = \frac{1}{3}$. $S = \sum_{x=0}^{\infty} (x+1) r^x$ Expanding the terms: $S = (0+1)r^0 + (1+1)r^1 + (2+1)r^2 + (3+1)r^3 + \ldots$ $S = 1 + 2r + 3r^2 + 4r^3 + \ldots \quad \ldots (1)$ This is an arithmetic-geometric series. To sum it, we use the "shift and subtract" method. Multiply equation (1) by $r$: $rS = \quad r + 2r^2 + 3r^3 + \ldots \quad \ldots (2)$ Subtract equation (2) from equation (1): $S - rS = (1 + 2r + 3r^2 + \ldots) - (r + 2r^2 + 3r^3 + \ldots)$ $S(1-r) = 1 + (2r - r) + (3r^2 - 2r^2) + (4r^3 - 3r^3) + \ldots$ $S(1-r) = 1 + r + r^2 + r^3 + \ldots$ The right-hand side is an infinite geometric series with first term $a=1$ and common ratio $r$. Since $r = \frac{1}{3}$ and $|r|<1$, its sum is $\frac{a}{1-r} = \frac{1}{1-r}$. $S(1-r) = \frac{1}{1-r}$ Now, solve for $S$: $S = \frac{1}{(1-r)^2}$ Substitute $r = \frac{1}{3}$: $S = \frac{1}{\left(1-\frac{1}{3}\right)^2} = \frac{1}{\left(\frac{2}{3}\right)^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4}$ Substitute the value of $S$ back into the equation for $k$: $k \cdot S = 1$ $k \cdot \frac{9}{4} = 1$ Solving for $k$: $k = \frac{4}{9}$

Step 3: Calculate $P(X \geq 3)$ Using the complement rule: $P(X \geq 3) = 1 - P(X < 3)$ Since $X$ is a discrete variable taking non-negative integer values, $P(X < 3)$ is the sum of probabilities for $X=0, X=1,$ and $X=2$: $P(X < 3) = P(X=0) + P(X=1) + P(X=2)$ Now, we calculate each term using $P(X=x) = k(x+1)3^{-x}$ with $k=\frac{4}{9}$:

For $x=0$: $P(X=0) = \frac{4}{9}(0+1)3^{-0} = \frac{4}{9}(1)(1) = \frac{4}{9}$

For $x=1$: $P(X=1) = \frac{4}{9}(1+1)3^{-1} = \frac{4}{9}(2)\left(\frac{1}{3}\right) = \frac{8}{27}$

For $x=2$: $P(X=2) = \frac{4}{9}(2+1)3^{-2} = \frac{4}{9}(3)\left(\frac{1}{9}\right) = \frac{12}{81} = \frac{4}{27}$

Now, sum these probabilities to find $P(X < 3)$: $P(X < 3) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27}$ To add these fractions, find a common denominator, which is 27: $P(X < 3) = \frac{4 \cdot 3}{9 \cdot 3} + \frac{8}{27} + \frac{4}{27} = \frac{12}{27} + \frac{8}{27} + \frac{4}{27}$ $P(X < 3) = \frac{12+8+4}{27} = \frac{24}{27}$ Simplify the fraction by dividing the numerator and denominator by 3: $P(X < 3) = \frac{8}{9}$ Finally, apply the complement rule: $P(X \geq 3) = 1 - P(X < 3) = 1 - \frac{8}{9} = \frac{1}{9}$

3. Common Mistakes & Tips

• Forgetting Total Probability: Always remember to use $\sum P(X=x)=1$ to find any unknown constants in the PMF. This is the starting point for most such problems.
• Errors in Summing Series: Be proficient with summing infinite series, especially geometric and arithmetic-geometric series. The "shift and subtract" method is a reliable technique for arithmetic-geometric series. Alternatively, you can recall the differentiation shortcut: if $\sum_{x=0}^{\infty} r^x = \frac{1}{1-r}$, then $\sum_{x=0}^{\infty} (x+1)r^x = \frac{d}{dr}\left(\sum_{x=0}^{\infty} r^{x+1}\right) = \frac{d}{dr}\left(\frac{r}{1-r}\right)$ or, more directly, $\sum_{x=0}^{\infty} (x+1)r^x = \sum_{y=1}^{\infty} yr^{y-1} = \frac{d}{dr}\left(\sum_{y=0}^{\infty} r^y\right) = \frac{d}{dr}\left(\frac{1}{1-r}\right) = \frac{1}{(1-r)^2}$.
• Ignoring the Complement Rule: For probabilities like $P(X \geq a)$ or $P(X > a)$ in infinite discrete distributions, the complement rule ($P(A) = 1 - P(A^c)$) simplifies the calculation from an infinite sum to a finite one, saving significant time and reducing error potential.

4. Summary

This problem effectively tests the understanding of discrete probability distributions. We first determined the constant $k$ by utilizing the fundamental property that the sum of all probabilities must equal 1. This involved summing an infinite arithmetic-geometric series. Subsequently, to calculate $P(X \geq 3)$, we strategically employed the complement rule, which transformed an infinite sum into a manageable finite sum of the first three probabilities ($P(X=0), P(X=1), P(X=2)$). The final result was obtained by subtracting this sum from 1.

The final answer is $\boxed{\frac{1}{9}}$, which corresponds to option (A).