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

Q: Wait, why do we use the complement rule instead of just summing from $x=3$ to infinity?

That is a brilliant question! While you *could* sum from $x=3$ to infinity, it would require calculating an infinite series again. The complement rule $P(X \ge 3) = 1 - P(X < 3)$ is a powerful shortcut. It reduces an infinite problem to a finite one: calculating only three simple terms ($x=0, 1, 2$). Always look for the path of least resistance in probability!

Q: How do I know for sure that the series is an Arithmetico-Geometric Progression (AGP)?

Look at the terms of the series: $(x+1)$ and $(1/3)^x$. The term $(x+1)$ generates the sequence $1, 2, 3, 4, \dots$, which is an Arithmetic Progression (AP) with first term $a=1$ and common difference $d=1$. The term $(1/3)^x$ generates $1, 1/3, 1/9, 1/27, \dots$, which is a Geometric Progression (GP) with common ratio $r=1/3$. Since each term is the product of an AP term and a GP term, it is by definition an AGP.

Q: What if the series didn't converge? Would the problem still be valid?

In the context of a probability mass function, the series *must* converge to a finite value for the distribution to be valid. If the sum of probabilities were infinite, the total probability could not be $1$, and the distribution would not exist. In JEE problems, you can trust that the given function is a valid probability distribution, meaning the series will always converge.

Question

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

Options:

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

(B) $\frac{4}{9}$

(C) $\frac{8}{27}$

(D) $\frac{1}{9}$

Sigma Percentile · Accepted Answer

### The Elegance of Infinite Series: A Journey into Probability Distributions

Welcome, future IITian! Today, we are going to unravel the mystery of a discrete probability distribution. It might look like a daunting infinite sum at first glance, but as we peel back the layers, you will see the beautiful symmetry hidden within.

Our journey begins with the probability mass function $P(X = x) = k(x + 1)3^{-x}$ for $x = 0, 1, 2, 3, \dots$. Our mission is to find $P(X \ge 3)$.

### Phase 1: The Hunt for the Constant $k$

Before we can calculate any specific probability, we must find the value of the constant $k$. In the world of probability, there is one golden rule: the total probability of all possible outcomes must equal $1$.

Mathematically, this is expressed as:

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

By substituting our function, we get:

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

Since $k$ is a constant, we can pull it out of the summation:

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

Now, we have a clear target: calculate the sum $S = \sum_{x=0}^{\infty} (x + 1) \left( \frac{1}{3} \right)^x$.

### Phase 2: Decoding the AGP

Let us expand this series to see its true nature:

$$S = 1 + 2\left( \frac{1}{3} \right) + 3\left( \frac{1}{3} \right)^2 + 4\left( \frac{1}{3} \right)^3 + \dots$$

The coefficients $1, 2, 3, 4, \dots$ form an Arithmetic Progression (AP) with $a=1$ and $d=1$. The terms $1, 1/3, 1/9, 1/27, \dots$ form a Geometric Progression (GP) with common ratio $r=1/3$. This is the classic Arithmetico-Geometric Progression (AGP).

We use the standard formula for the sum of an infinite AGP:

$$S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$$

Plugging in our values, $a=1$, $d=1$, and $r=1/3$, we get:

$$S = \frac{1}{1 - 1/3} + \frac{1 \cdot (1/3)}{(1 - 1/3)^2}$$

Simplifying this:

$$S = \frac{1}{2/3} + \frac{1/3}{4/9} = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$$

With $S = 9/4$, our normalization equation becomes $k(9/4) = 1$, which gives us **$k = 4/9$**. We have successfully unlocked the constant!

### Phase 3: The Power of the Complement

Now, we need to find $P(X \ge 3)$. Instead of summing from $x=3$ to infinity, we use the complement rule:

$$P(X \ge 3) = 1 - P(X < 3)$$

This is much simpler! We only need to calculate $P(X=0) + P(X=1) + P(X=2)$.

Using $k=4/9$:

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

$$P(X=1) = k(2)\left( \frac{1}{3} \right) = \frac{8}{27}$$

$$P(X=2) = k(3)\left( \frac{1}{9} \right) = \frac{12}{81} = \frac{4}{27}$$

Summing these:

$$P(X < 3) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{12}{27} + \frac{8}{27} + \frac{4}{27} = \frac{24}{27} = \frac{8}{9}$$

Finally, the result is:

$$P(X \ge 3) = 1 - \frac{8}{9} = \mathbf{\frac{1}{9}}$$

The elegance of the result is truly satisfying. You have navigated the infinite, identified the pattern, and used the complement rule to reach the finish line. Keep this analytical mindset, and no JEE problem will ever stand in your way!

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

Select Answer:

Visualized Solution

The Probability Distribution

The Total Probability Law

Setting Up the Infinite Sum

Expanding the Series

Identifying the AGP

AGP Sum Formula

Calculating the Sum $S$

Finding the Constant $k$

The Target Probability

The Complement Rule

Calculating $P (X < 3)$

Simplifying $P (X < 3)$

Final Answer

The Sigma Insight: Random Variables and Probability Distributions

The Elegance of Infinite Series

Phase 1

Phase 2

Phase 3

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