Let X be a random variable such that the probability function of a distribution is given by P(X = 0)

Q: Wait, why does the $P(X=0)$ term disappear in the mean calculation?

Great observation! The definition of the mean is $E(X) = \sum x_i p_i$. When $x_i = 0$, the product $0 \cdot P(X=0)$ is simply $0$. It contributes nothing to the expected value, so we focus our energy on the infinite series starting from $j=1$.

Q: Is it always valid to subtract two infinite series like that?

Yes, provided the series is absolutely convergent. Since our series $\sum j/3^j$ converges (which you can verify via the ratio test), the algebraic manipulation of shifting and subtracting is perfectly rigorous. Think of it as 'telescoping' the complexity away!

Q: Why is the common ratio for the even probability $1/9$ and not $1/3$?

Because we are skipping terms! We only want $j=2, 4, 6, \dots$. The sequence of probabilities is $1/3^2, 1/3^4, 1/3^6, \dots$. To get from $1/3^2$ to $1/3^4$, you must multiply by $1/3^2$, which is $1/9$. Always identify the first term and the ratio of the *specific* sequence you are summing.

Q: What if the sum didn't start at $j=1$? Would the formula change?

The logic remains identical! The key is to identify the first term $a$ and the common ratio $r$. If the series started at $j=2$, your first term would simply be the value at $j=2$. The 'shift-and-subtract' method is robust enough to handle any starting index.

Q: Is there a shortcut for the mean of this distribution?

There is! This distribution is related to a Geometric Distribution. While the AGP method is the standard JEE approach, recognizing the structure of $\sum j r^j$ as the derivative of a geometric series is a powerful 'pro' move for competitive exams.

Question

Let $X$ be a random variable such that the probability function of a distribution is given by $P(X = 0) = 1/2, P(X = j) = 1/3^j (j = 1, 2, 3, \dots, \infty)$. Then the mean of the distribution and $P(X 	ext{ is positive and even})$ respectively are:

Options:

(A) $3/8$ and $1/8$

(B) $3/4$ and $1/8$

(C) $3/4$ and $1/9$

(D) $3/4$ and $1/16$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we aren't just solving a probability problem; we are peeling back the layers of an infinite distribution.

We are given $P(X=0) = 1/2$, and for every positive integer $j$, $P(X=j) = 1/3^j$. It looks intimidating, but in the world of mathematics, infinity is a playground.

### The Anatomy of the Mean

To find the mean $E(X)$, we use the fundamental definition:

$$E(X) = \sum_{j=0}^{\infty} j \cdot P(X=j)$$

When we plug in our values, the first term $0 \cdot P(X=0)$ vanishes. We are left with the series:

$$E(X) = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \dots$$

This is a classic **Arithmetico-Geometric Progression (AGP)**. It is a hybrid where the numerator grows linearly while the denominator grows exponentially.

### The Shift-and-Subtract Strategy

To tame this series, let $S = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \dots$. If we multiply this entire sum by the common ratio $r = 1/3$, we get:

$$\frac{1}{3}S = \frac{1}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \dots$$

Now, subtract $\frac{1}{3}S$ from $S$ so that the terms align perfectly:

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

This simplifies to:

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

The complexity collapses into a simple infinite Geometric Progression with first term $a = 1/3$ and ratio $r = 1/3$. Using the sum formula $S = \frac{a}{1-r}$, we find:

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

Solving for $S$, we obtain the mean:

**$E(X) = 3/4$**

### The Elegance of Even Probabilities

Now, we tackle the probability that $X$ is positive and even. We are looking for $P(X=2) + P(X=4) + P(X=6) + \dots$.

Substituting our probability function, this becomes:

$$P(X 	ext{ is positive and even}) = \frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} + \dots$$

This is a pure Geometric Progression where the first term $a = 1/9$ and the common ratio $r = 1/9$. Applying the sum formula:

$$Sum = \frac{1/9}{1 - 1/9} = \frac{1/9}{8/9} = \frac{1}{8}$$

The infinite series yields to our persistence. We have found the mean to be **$3/4$** and the probability of a positive even outcome to be **$1/8$**.

Animated Solution for Mathematics - Probability: Let $X$ be a random variable such that the probability function of a distribution is given by $P (X = 0) = 1/2, P (X = j) = 1/ 3^{j} (j = 1, 2, 3, \dots, \infty)$ . Then the mean of the distribution and $P (X is positive and even)$ respectively are:

Select Answer:

Visualized Solution

Understanding the Distribution

Defining the Mean $E (X)$

Expanding the Series

Solving the AGP (Part 1)

Solving the AGP (Part 2)

Calculating the Mean Value

Probability of Positive Even $X$

Solving the Even Probability GP

Final Calculation

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Anatomy of the Mean

The Shift-and-Subtract Strategy

The Elegance of Even Probabilities

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