Let X be a random variable, and let P(X = x) denote the probability that X takes the values x. Suppo

Q: Wait, why can we assume $P(X=x) = mx+c$?

The problem explicitly states that the points $(x, P(X=x))$ lie on a fixed straight line. In coordinate geometry, any set of points on a line satisfies the linear equation $y = mx + c$. Here, $y$ is the probability $P(X=x)$, so the linear relationship is a direct consequence of the problem's geometric constraint.

Q: Is there a faster way to find $m$ and $c$ using symmetry?

You have a sharp eye! Because the mean is $2.5$, which is exactly the midpoint of the range $[0, 4]$, the distribution is symmetric. This implies the probabilities are symmetric about $x=2$. While you could use this to simplify the algebra, solving the system of equations is the most robust and foolproof method for the JEE.

Q: Why did we use $Var(X) = E[X^2] - (E[X])^2$ instead of the definition?

You could use the definition $Var(X) = \sum (x - E[X])^2 P(X=x)$, but it requires calculating the deviation from the mean for every single point. The computational formula $Var(X) = E[X^2] - (E[X])^2$ is generally faster and less prone to arithmetic errors, especially when you have already calculated $E[X]$.

Question

Let $X$ be a random variable, and let $P(X = x)$ denote the probability that $X$ takes the values $x$. Suppose that the points $(x, P(X = x))$, $x = 0, 1, 2, 3, 4$, lie on a fixed straight line in the $xy$-plane, and $P(X = x) = 0$ for all $x \in \mathbb{R} - \{0, 1, 2, 3, 4\}$. If the mean of $X$ is $\frac{5}{2}$ and the variance of $X$ is $\alpha$, then the value of $24\alpha$ is ________.

Sigma Percentile · Accepted Answer

### The Geometry of Chance: A Journey into Linear Probability

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are uncovering the hidden geometry within a probability distribution.

Often, we treat probability and coordinate geometry as separate silos, but this problem forces them to collide in the most elegant way possible.

### Analyzing the Setup

Imagine you are plotting the probabilities of a random variable $X$ that takes values in the set $\{0, 1, 2, 3, 4\}$. The problem states that the points $(x, P(X=x))$ lie on a straight line.

If the points are collinear, the probability function must be a linear function of $x$. We define this as:

$$P(X=x) = mx + c$$

Here, $m$ is the slope of our probability line, and $c$ is the y-intercept. Our mission is to find these two constants, $m$ and $c$, to unlock the entire distribution.

### The Constraints of Reality

In the world of probability, there are two fundamental laws that govern everything. First, the sum of all probabilities must be exactly $1$:

$$\sum_{x=0}^{4} P(X=x) = 1$$

Substituting our linear form, we get:

$$\sum_{x=0}^{4} (mx + c) = 1$$

Expanding this, we have $m(0 + 1 + 2 + 3 + 4) + 5c = 1$, which simplifies to the anchor equation:

$$10m + 5c = 1$$

### The Center of Gravity

Next, we use the mean of the distribution, which is given as $E[X] = \frac{5}{2}$. The mean is essentially the center of mass of our probability distribution.

The formula is $E[X] = \sum_{x=0}^{4} x \cdot P(X=x)$. Substituting our linear function, we get:

$$\sum_{x=0}^{4} x(mx + c) = \frac{5}{2} \Rightarrow \sum_{x=0}^{4} (mx^2 + cx) = \frac{5}{2}$$

Evaluating the sums of squares and integers, we get $m(30) + c(10) = 2.5$, or:

$$30m + 10c = 2.5$$

Now, we have a system of two equations: $10m + 5c = 1$ and $30m + 10c = 2.5$. Solving this system, we find:

$$m = \frac{1}{20}, \quad c = \frac{1}{10}$$

### The Variance

With $m$ and $c$ in hand, we have fully defined our distribution. The final hurdle is finding the variance, $\alpha$.

We use the standard identity $Var(X) = E[X^2] - (E[X])^2$. We already know $E[X] = 2.5$, so we just need $E[X^2]$. This is calculated as:

$$E[X^2] = \sum_{x=0}^{4} x^2 \left( \frac{1}{20}x + \frac{1}{10} ight) = \frac{1}{20}\sum_{x=0}^{4} x^3 + \frac{1}{10}\sum_{x=0}^{4} x^2$$

Using the sum of cubes ($100$) and sum of squares ($30$), we get:

$$E[X^2] = \frac{100}{20} + \frac{30}{10} = 5 + 3 = 8$$

Finally, the variance is:

$$\alpha = 8 - (2.5)^2 = 8 - 6.25 = 1.75 = \frac{7}{4}$$

### Final Calculation

The question asks for $24\alpha$. Substituting our value:

$$24 	imes \frac{7}{4} = 6 	imes 7 = 42$$

We have successfully navigated the intersection of geometry and probability. The final answer is **42**.

Enter Numerical Value:

Visualized Solution

Visualizing the Probability Distribution

Defining the Linear Equation

Law of Total Probability

Forming Equation 1

Using the Mean Condition

Setting up the Mean Equation

Evaluating the Sum of Squares

Solving for $m$ and $c$

The Variance Formula

Setting up $E [X^{2}]$

Evaluating $E [X^{2}]$

Calculating Variance $α$

Final Calculation

The Sigma Insight: Random Variables and Probability Distributions

The Geometry of Chance

Analyzing the Setup

The Constraints of Reality

The Center of Gravity

The Variance

Final Calculation

Similar Questions

Let $X$ be a random variable with distribution. $x P (X = x) - 2 \frac{1}{5} - 1 a 3 \frac{1}{3} 4 \frac{1}{5} 6 b$ If the mean of $X$ is 2.3 and variance of $X$ is $σ^{2}$ , then $100 σ^{2}$ is equal to :

If the mean of the following probability distribution of a random variable $X$ : $X P (X) 0 a 2 2 a 4 a + b 6 2 b 8 3 b$ is $46/9$ , then the variance of the distribution is

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Let $A = [a_{ij}]$ be a $2 \times 2$ matrix such that $a_{ij} \in {0, 1}$ for all i and j. Let the random variable X denote the possible values of the determinant of the matrix A. Then, the variance of X is :

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From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $σ^{2}$ , then $96 σ^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Probability Distribution

Defining the Linear Equation

Law of Total Probability

Forming Equation 1

Using the Mean Condition

Setting up the Mean Equation

Evaluating the Sum of Squares

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

The Variance Formula

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E[X2]

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E[X2]

Calculating Variance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Final Calculation

The Sigma Insight: Random Variables and Probability Distributions

The Geometry of Chance

Analyzing the Setup

The Constraints of Reality

The Center of Gravity

The Variance

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean of the following probability distribution of a random variable X: XP(X)​0a​22a​4a+b​62b​83b​​ is 46/9, then the variance of the distribution is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ and σ2 denote the mean and variance of X, then the value of 64(μ+σ2) is:

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Setting up $E [X^{2}]$

Evaluating $E [X^{2}]$

Calculating Variance $α$

If the mean of the following probability distribution of a random variable $X$ : $X P (X) 0 a 2 2 a 4 a + b 6 2 b 8 3 b$ is $46/9$ , then the variance of the distribution is

A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is: