Key Concepts and Formulas

1. Normalization Property of Probability Distributions: For any discrete random variable, the sum of probabilities of all possible outcomes must be equal to 1. This is a fundamental property for a valid probability distribution: $\sum_{\text{all possible } x} \mathrm{P}(\mathrm{X}=x) = 1$
2. Sum of an Infinite Arithmetico-Geometric Progression (AGP): An infinite series where each term is the product of a term from an Arithmetic Progression (AP) and a term from a Geometric Progression (GP). A specific form relevant here is $\sum_{n=0}^{\infty} (n+1)r^n = \frac{1}{(1-r)^2}$, provided that the common ratio of the GP, $r$, satisfies $|r|<1$.
3. Complementary Probability: The probability of an event occurring is 1 minus the probability of its complementary event. This is often useful for calculating probabilities of "at least" or "greater than or equal to" events: $\mathrm{P}(A) = 1 - \mathrm{P}(A^c)$

Step-by-Step Solution

Step 1: Determine the constant $k$ using the Normalization Property.

• What we are doing: Our first goal is to find the value of the unknown constant $k$ in the given probability mass function (PMF).
• Why this is needed: A probability distribution must sum to 1 over all possible outcomes. This allows us to set up an equation to solve for $k$.
• Mathematical Derivation: The random variable $\mathrm{X}$ takes values $x=0, 1, 2, 3, \ldots$. The PMF is given by $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1) 3^{-x}$. Applying the normalization property: $\sum_{x=0}^{\infty} \mathrm{P}(\mathrm{X}=x) = 1$ Substitute the given PMF: $\sum_{x=0}^{\infty} \mathrm{k}(x+1) 3^{-x} = 1$ Since $k$ is a constant, we can factor it out of the summation: $k \sum_{x=0}^{\infty} (x+1) \left(\frac{1}{3}\right)^x = 1$ Let's denote the infinite sum as $S$: $S = \sum_{x=0}^{\infty} (x+1) \left(\frac{1}{3}\right)^x$ This is an infinite Arithmetico-Geometric Progression (AGP). The terms are: For $x=0$: $(0+1)(1/3)^0 = 1 \cdot 1 = 1$ For $x=1$: $(1+1)(1/3)^1 = 2 \cdot (1/3) = 2/3$ For $x=2$: $(2+1)(1/3)^2 = 3 \cdot (1/9) = 3/9$ And so on: $S = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \ldots$ This series matches the general form $\sum_{n=0}^{\infty} (n+1)r^n$ with $r = 1/3$. Since $|r|=|1/3|<1$, the sum converges and can be calculated using the formula: $S = \frac{1}{(1-r)^2}$ Substitute $r=1/3$: $S = \frac{1}{\left(1 - \frac{1}{3}\right)^2} = \frac{1}{\left(\frac{2}{3}\right)^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4}$ Now, substitute the value of $S$ back into the equation for $k$: $k \cdot \frac{9}{4} = 1$ Solving for $k$: $k = \frac{4}{9}$

Step 2: Calculate $\mathrm{P}(\mathrm{X} \geq 2)$ using Complementary Probability.

• What we are doing: We need to find the probability that $\mathrm{X}$ takes a value greater than or equal to 2.
• Why this is needed: Directly summing $\mathrm{P}(\mathrm{X}=2) + \mathrm{P}(\mathrm{X}=3) + \ldots$ involves another infinite sum, which can be more complex. Using the complementary probability simplifies the calculation significantly.
• Mathematical Derivation: The event $\mathrm{X} \geq 2$ includes all outcomes where $\mathrm{X}$ is $2, 3, 4, \ldots$. The complementary event, $\mathrm{X} < 2$, includes outcomes where $\mathrm{X}$ is $0, 1$ (since $\mathrm{X}$ can only take non-negative integer values). Using the complementary probability formula: $\mathrm{P}(\mathrm{X} \geq 2) = 1 - \mathrm{P}(\mathrm{X} < 2)$ $\mathrm{P}(\mathrm{X} \geq 2) = 1 - (\mathrm{P}(\mathrm{X}=0) + \mathrm{P}(\mathrm{X}=1))$ Now we calculate $\mathrm{P}(\mathrm{X}=0)$ and $\mathrm{P}(\mathrm{X}=1)$ using the value of $k = 4/9$ that we found. Recall the PMF: $\mathrm{P}(\mathrm{X}=x) = k(x+1)3^{-x}$. For $\mathrm{X}=0$: $\mathrm{P}(\mathrm{X}=0) = \frac{4}{9}(0+1)3^{-0} = \frac{4}{9}(1)(1) = \frac{4}{9}$ For $\mathrm{X}=1$: $\mathrm{P}(\mathrm{X}=1) = \frac{4}{9}(1+1)3^{-1} = \frac{4}{9}(2)\left(\frac{1}{3}\right) = \frac{8}{27}$ Substitute these probabilities back into the complementary probability formula: $\mathrm{P}(\mathrm{X} \geq 2) = 1 - \left(\frac{4}{9} + \frac{8}{27}\right)$ To add the fractions, find a common denominator, which is 27: $\mathrm{P}(\mathrm{X} \geq 2) = 1 - \left(\frac{4 \times 3}{9 \times 3} + \frac{8}{27}\right) = 1 - \left(\frac{12}{27} + \frac{8}{27}\right)$ $\mathrm{P}(\mathrm{X} \geq 2) = 1 - \frac{20}{27}$ $\mathrm{P}(\mathrm{X} \geq 2) = \frac{27}{27} - \frac{20}{27} = \frac{7}{27}$

Common Mistakes & Tips

• Starting Index of Summation: Always pay close attention to the specified range of $x$. Here, $x=0, 1, 2, \ldots$. A common mistake is to start the summation from $x=1$ instead of $x=0$, which would lead to an incorrect value of $k$.
• AGP Sum Formula: Ensure you use the correct formula for the sum of the AGP. For $\sum_{n=0}^{\infty} (n+1)r^n$, the sum is $\frac{1}{(1-r)^2}$. There are variations for different starting terms or forms of the arithmetic progression.
• Fraction Arithmetic: Be careful with adding and subtracting fractions, especially when finding common denominators. Errors here can easily lead to an incorrect final answer.

Summary This problem required a two-step approach. First, we used the normalization property of probability distributions, summing the given probability mass function over all possible values of $x$ (from 0 to infinity) and equating it to 1. This sum was identified as an infinite Arithmetico-Geometric Progression (AGP), allowing us to solve for the constant $k$. Second, to find $\mathrm{P}(\mathrm{X} \geq 2)$, we efficiently used the complementary probability rule, calculating $1 - (\mathrm{P}(\mathrm{X}=0) + \mathrm{P}(\mathrm{X}=1))$ instead of an infinite sum. The final calculated probability is $\frac{7}{27}$.

The final answer is $\boxed{\frac{7}{27}}$, which corresponds to option (C).