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JEE Main · Mathematics · Probability & Statistics
MCQ · Mathematics · Random Variable
Q. 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:
A12
B60 ✓
C30
D3
✅ Correct Answer: (B) 60
Step-by-Step Solution
1
Simplify the Given Condition
We know that $\sigma^2 = E(X^2) – [E(X)]^2$.
Also, $\mu = E(X)$.
Substituting these into $\sigma^2 + \mu^2 = 2$:
$$(E(X^2) – \mu^2) + \mu^2 = 2$$
$$\implies E(X^2) = 2$$
2
Sum of Probabilities = 1
The total probability must be 1:
$$\frac{2a+1}{30} + \frac{8a-1}{30} + \frac{4a+1}{30} + b = 1$$
$$\frac{14a + 1}{30} + b = 1 \implies b = 1 – \frac{14a + 1}{30} = \frac{29 – 14a}{30} \quad \dots \text{(i)}$$
3
Apply $E(X^2) = 2$
$E(X^2) = \sum x_i^2 \cdot P(X=x_i)$:
$$0^2\left(\frac{2a+1}{30}\right) + 1^2\left(\frac{8a-1}{30}\right) + 2^2\left(\frac{4a+1}{30}\right) + 3^2(b) = 2$$
$$\frac{8a-1}{30} + \frac{16a+4}{30} + 9b = 2$$
$$\frac{24a+3}{30} + 9b = 2$$
4
Solve for $a$
Substitute $b$ from eq (i):
$$\frac{24a+3}{30} + 9\left(\frac{29-14a}{30}\right) = 2$$
$$24a + 3 + 261 – 126a = 60$$
$$-102a + 264 = 60 \implies 102a = 204$$
$$a = 2$$
5
Final Ratio $\frac{a}{b}$
Using $a=2$ in eq (i):
$b = \frac{29 – 14(2)}{30} = \frac{29 – 28}{30} = \frac{1}{30}$
$\frac{a}{b} = \frac{2}{1/30} = 2 \times 30 = 60$
✅ Correct Option: (B)
$\frac{a}{b} = \frac{2}{1/30} = 2 \times 30 = 60$
Related Theory: Probability Distributions
📌 Understanding Random Variables
A Random Variable is a real-valued function defined over a sample space. In discrete distributions, like this problem, the variable $X$ takes specific values with associated probabilities $P(X=x_i)$. Two fundamental constraints always apply:
1. $P(x_i) \geq 0$ for all $i$.
2. $\sum P(x_i) = 1$ (Total probability axiom).
1. $P(x_i) \geq 0$ for all $i$.
2. $\sum P(x_i) = 1$ (Total probability axiom).
📌 Mean, Variance, and Moments
The Mean ($\mu$) or Expectation $E(X)$ represents the weighted average of the outcomes.
The Variance ($\sigma^2$) measures the spread of the distribution.
Key Identities used in JEE: $E(X) = \sum x_i p_i$ $E(X^2) = \sum x_i^2 p_i$ $\sigma^2 = E(X^2) – [E(X)]^2$ In many JEE questions, expressions like $\sigma^2 + \mu^2$ are given to simplify the problem directly to $E(X^2)$, bypassing the need to calculate the mean separately.
The Variance ($\sigma^2$) measures the spread of the distribution.
Key Identities used in JEE: $E(X) = \sum x_i p_i$ $E(X^2) = \sum x_i^2 p_i$ $\sigma^2 = E(X^2) – [E(X)]^2$ In many JEE questions, expressions like $\sigma^2 + \mu^2$ are given to simplify the problem directly to $E(X^2)$, bypassing the need to calculate the mean separately.
📌 Statistical Parameters Calculation
When solving for unknown parameters ($a, b$) in a distribution:
• Step 1: Use $\sum P = 1$ to create the first linear equation.
• Step 2: Use the given statistical property (Mean, Variance, or $E(X^2)$) to create the second equation.
• Step 3: Solve the system of equations. Always check if probabilities remain positive ($P \geq 0$) after finding the constants.
• Step 1: Use $\sum P = 1$ to create the first linear equation.
• Step 2: Use the given statistical property (Mean, Variance, or $E(X^2)$) to create the second equation.
• Step 3: Solve the system of equations. Always check if probabilities remain positive ($P \geq 0$) after finding the constants.
📌 Why $\sigma^2 + \mu^2 = E(X^2)$?
This is a common property derived from the definition of variance:
$$\sigma^2 = \text{Mean of squares} – \text{Square of mean}$$
$$\sigma^2 = E(X^2) – \mu^2$$
Adding $\mu^2$ to both sides yields $E(X^2) = \sigma^2 + \mu^2$. This “shortcut” is frequently tested in JEE Main to see if students can avoid long, tedious calculations.
📌 Discrete vs Continuous Variance
In Discrete cases (like this one), we use summations ($\sum$). In Continuous cases, we use integrals ($\int$). However, the relationship $\sigma^2 = E(X^2) – \mu^2$ remains universal across all probability distributions.
📌 Key Result Summary
Total Prob = 1
$E(X^2) = 2$
$a = 2$
$b = 1/30$
$a/b = 60$
Frequently Asked Questions
1. Why did we set $E(X^2) = 2$?
Because the formula for variance is $\sigma^2 = E(X^2) – \mu^2$. Adding $\mu^2$ gives $\sigma^2 + \mu^2 = E(X^2)$, which was given as 2.
2. How was the value of $b$ calculated?
By ensuring the sum of all probabilities $(P_0 + P_1 + P_2 + P_3)$ equals 1.
3. Can $a$ or $b$ be negative?
In a valid probability distribution, each individual probability $P(X=x)$ must be $\geq 0$. Here $a=2$ and $b=1/30$ satisfy this.
4. What is the mean of this distribution?
Using $a=2$, $\mu = E(X) = 1(15/30) + 2(9/30) + 3(1/30) = 36/30 = 1.2$.
5. What is the variance?
$\sigma^2 = E(X^2) – \mu^2 = 2 – (1.2)^2 = 2 – 1.44 = 0.56$.
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