A random variable X takes values 0, 1, 2, 3 with probabilities (2a+1)/30, (8a-1)/30, (4a+1)/30, b re

Q: Why does $\sigma^2 + \mu^2$ simplify to $E(X^2)$? Isn't that just a definition?

It is indeed a definition, but it is a powerful one! In JEE problems, recognizing this identity is the difference between a 2-minute solution and a 10-minute algebraic nightmare. It allows us to bypass the calculation of the mean $\mu$ entirely.

Q: Do I really need to calculate $\mu$?

Absolutely not! That is the beauty of the identity $\sigma^2 + \mu^2 = E(X^2)$. By recognizing this, you bypass the tedious calculation of the mean $\mu$ entirely, saving precious time for other problems.

Q: How do I know if my values for $a$ and $b$ are correct?

Always perform a sanity check! Each probability $P(X=x_i)$ must lie in the interval $[0, 1]$. If you calculate a value that makes a probability negative or greater than 1, you know your algebra has gone astray.

Question

A random variable $X$ takes values $0, 1, 2, 3$ with probabilities $\frac{2a+1}{30}, \frac{8a-1}{30}, \frac{4a+1}{30}, b$ respectively, where $a, b \in \mathbb{R}$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^2 + \mu^2 = 2$. Then $\frac{a}{b}$ is equal to :

Options:

(A) 12

(B) 30

(C) 60

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The random variable $X$ takes values $0, 1, 2, 3$ with probabilities $P(0) = \frac{2a+1}{30}$, $P(1) = \frac{8a-1}{30}$, $P(2) = \frac{4a+1}{30}$, and $P(3) = b$.

The fundamental law of probability dictates that the sum of all probabilities must equal $1$. We express this as:

$$\frac{2a+1}{30} + \frac{8a-1}{30} + \frac{4a+1}{30} + b = 1$$

Combining the terms with the common denominator of $30$, we obtain:

$$\frac{14a + 1}{30} + b = 1 \implies b = \frac{29 - 14a}{30}$$

This equation successfully links our two unknowns, $a$ and $b$.

### The Shortcut to Genius

We are given the condition $\sigma^2 + \mu^2 = 2$. Many students rush to calculate the mean $\mu = E(X)$ and the variance $\sigma^2 = E(X^2) - \mu^2$.

However, by substituting the definition of variance into the given condition, we observe a significant simplification:

$$\sigma^2 + \mu^2 = (E(X^2) - \mu^2) + \mu^2 = E(X^2) = 2$$

This is the **"aha!" moment**. We do not need to calculate the mean; we only need the expected value of the square of the random variable, $E(X^2)$.

### The Final Calculation

We define $E(X^2) = \sum x_i^2 P(x_i)$. Substituting our specific values, we get:

$$0^2 \cdot \left( \frac{2a+1}{30} \right) + 1^2 \cdot \left( \frac{8a-1}{30} \right) + 2^2 \cdot \left( \frac{4a+1}{30} \right) + 3^2 \cdot b = 2$$

This simplifies to the following linear equation:

$$\frac{8a-1}{30} + 4 \cdot \left( \frac{4a+1}{30} \right) + 9b = 2$$

Substituting our expression for $b = \frac{29 - 14a}{30}$ into the equation, the entire expression collapses into a single variable $a$:

$$\frac{8a - 1 + 16a + 4}{30} + 9 \left( \frac{29 - 14a}{30} \right) = 2$$

Multiplying by $30$ to clear the denominator:

$$(24a + 3) + (261 - 126a) = 60$$

$$-102a + 264 = 60 \implies 102a = 204 \implies a = 2$$

With $a = 2$ determined, finding $b$ is trivial:

$$b = \frac{29 - 14(2)}{30} = \frac{1}{30}$$

Finally, the required ratio is:

$$\frac{a}{b} = \frac{2}{1/30} = \mathbf{60}$$

Select Answer:

Visualized Solution

$X$ and its Probability Distribution

Sum of Probabilities

Express $b$ in terms of $a$

Variance-Mean Relation

Formula for $E (X^{2})$

Substitute values into $E (X^{2})$

Substitute $b$ and simplify

Solve for $a$

Calculate $b$

Final Ratio $\frac{a}{b}$

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Shortcut to Genius

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X and its Probability Distribution

Sum of Probabilities

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Variance-Mean Relation

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Substitute values into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E(X2)

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Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

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

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