If the probability that the random variable X takes values x is given by P(X = x) = k(x + 1)3⁻ˣ, x =

Q: Wait, why do we use the complement rule? Can't we just sum from $x=2$ to infinity?

You absolutely could! But in the JEE, time is your most precious currency. Summing from $x=2$ to $\infty$ involves calculating an infinite series again. Using the complement rule reduces the problem to evaluating just two simple terms. Always look for the path of least resistance.

Q: How do I identify an AGP in the exam hall?

Look for the 'product' structure. If you see a term that looks like a linear polynomial (like $x+1$) multiplied by a geometric term (like $3^{-x}$), that is the signature of an AGP. It is a beautiful pattern waiting to be unraveled.

Q: Is it possible for the sum of the series to not equal 1?

In the context of a valid probability mass function, no. The definition of a PMF requires the total probability to be 1. If your sum doesn't equal 1, check your algebra—you likely missed a term or made a sign error in the AGP subtraction.

Question

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

Options:

(A) $\frac{7}{27}$

(B) $\frac{7}{18}$

(C) $\frac{11}{18}$

(D) $\frac{20}{27}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to peel back the layers of a problem that might look like a simple probability question at first glance, but is actually a beautiful dance of series and sequences.

When you see a probability mass function (PMF) defined as $P(X = x) = k(x + 1)3^{-x}$, do not panic. Instead, see it as a puzzle.

The universe of probability has one ironclad law: the sum of all possible outcomes must equal exactly one. This is our anchor. Our first mission is to find the constant $k$ by enforcing the condition:

$$\sum_{x=0}^{\infty} P(X = x) = 1$$

### The Arithmetico-Geometric Progression (AGP)

As we expand this summation, we get:

$$k \left[ (0+1)3^{0} + (1+1)3^{-1} + (2+1)3^{-2} + \dots \right] = 1$$

Let us look at the series inside the bracket: $1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \dots$. The numerators $1, 2, 3, 4, \dots$ are marching forward in an Arithmetic Progression with a common difference of $1$.

The denominators $1, 3, 9, 27, \dots$ are growing in a Geometric Progression with a common ratio of $\frac{1}{3}$. This is the Arithmetico-Geometric Progression (AGP), a classic JEE favorite.

To solve this, we use the 'Method of Differences'. We define our sum as $S = 1 + \frac{2}{3} + \frac{3}{9} + \dots$. Now, we multiply this entire series by the common ratio, $\frac{1}{3}$, to get:

$$\frac{1}{3}S = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \dots$$

When we subtract $\frac{1}{3}S$ from $S$, we align the terms with the same denominators. The arithmetic part vanishes, leaving us with a simple infinite geometric series:

$$S - \frac{1}{3}S = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$$

### The Final Calculation

The right side is now a standard infinite GP with first term $a=1$ and common ratio $r=\frac{1}{3}$. The sum formula $\frac{a}{1-r}$ gives us:

$$\frac{1}{1 - 1/3} = \frac{3}{2}$$

So, $\frac{2}{3}S = \frac{3}{2}$, which means $S = \frac{9}{4}$. Since $k \cdot S = 1$, we immediately find that **$k = \frac{4}{9}$**.

Now, we reach the final act: finding $P(X \ge 2)$. Instead of summing to infinity, we use the complement rule:

$$P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)]$$

Substituting our values, $P(X=0) = \frac{4}{9}(1) = \frac{4}{9}$ and $P(X=1) = \frac{4}{9}(2)(\frac{1}{3}) = \frac{8}{27}$. Adding these gives:

$$\frac{12}{27} + \frac{8}{27} = \frac{20}{27}$$

Finally, $1 - \frac{20}{27} = \frac{7}{27}$. The final answer is **$\frac{7}{27}$**.

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

Select Answer:

Visualized Solution

The Total Probability Principle

Expanding the Summation

Identifying the AGP Structure

The Method of Differences

Summing the Infinite GP

Determining the Constant $k$

Using the Complement Rule

Calculating $P (X = 0)$ and $P (X = 1)$

Final Result for $P (X \geq 2)$

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Arithmetico-Geometric Progression (AGP)

The Final Calculation

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Select Answer:

Visualized Solution

The Total Probability Principle

Expanding the Summation

Identifying the AGP Structure

The Method of Differences

Summing the Infinite GP

Determining the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Using the Complement Rule

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=0) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=1)

Final Result for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X≥2)

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Arithmetico-Geometric Progression (AGP)

The Final Calculation

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