Intermediate

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Random Variables & Probability Distributions

Master Random Variables and Probability Distributions for JEE Main and Board Exams. Covers discrete vs continuous random variables, probability mass function PMF conditions sum equals 1 all non-negative, cumulative distribution function CDF, expected value E(X)=Σxᵢpᵢ properties linearity E(aX+b), variance Var(X)=E(X²)-[E(X)]² properties Var(aX+b)=a²Var(X), standard deviation, finding k from sum condition, die coin toss distribution examples fully verified with board-level practice questions.

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A Random Variable assigns a numerical value to each outcome of a random experiment — it is the bridge between the abstract language of probability and the concrete world of numbers. Once we have a random variable, we can study its probability distribution (how probabilities are spread across its values), compute its expected value (long-run average), and measure its variance (spread). These are the tools needed for any quantitative analysis of uncertainty. In JEE and CBSE, problems on discrete random variables — constructing the probability distribution from a scenario, finding the mean and variance, and verifying that a given distribution is valid — are among the most directly scored questions in the chapter.

1. Random Variable — Definition and Types

A Random Variable (RV) X is a real-valued function defined on the sample space S: X : S → ℝ. It assigns a numerical value to each outcome.

Type	Description	Example
Discrete RV	Takes countable (finite or countably infinite) values	X = number of heads in 3 coin tosses (0,1,2,3)
Continuous RV	Takes any value in an interval	X = height of a randomly chosen person

CBSE and JEE syllabus covers discrete RVs only.

2. Probability Distribution (PMF)

The Probability Mass Function (PMF) of a discrete RV X is the table P(X = xᵢ) = pᵢ for each value xᵢ.

Conditions for a Valid Probability Distribution

$\sum_{i} p_{i} = 1 and p_{i} \geq 0 for all i$

Example — Tossing 2 Coins, X = Number of Heads

X	0	1	2	Sum
P(X)	1/4	2/4 = 1/2	1/4	1 ✓

Cumulative Distribution Function (CDF)

F(x) = P(X ≤ x) = sum of all P(X = xᵢ) for xᵢ ≤ x. CDF is non-decreasing; F(−∞) = 0; F(+∞) = 1.

3. Mean (Expected Value) of a Random Variable

The mean (expectation) of X is the long-run average value:

$μ = E (X) = \sum_{i} x_{i} \cdot p_{i}$

Properties of Expectation

Property	Formula
Constant	E(c) = c
Linear	E(aX + b) = aE(X) + b
Sum	E(X + Y) = E(X) + E(Y) (always, even if dependent)
Product (independent)	E(XY) = E(X)·E(Y)

For 2 coins: E(X) = 0×(1/4) + 1×(1/2) + 2×(1/4) = 0 + 1/2 + 1/2 = 1

4. Variance and Standard Deviation

$Var (X) = E (X^{2}) - [E (X)]^{2} = \sum x_{i}^{2} p_{i} - μ^{2}$

Standard Deviation: σ = √Var(X)

Properties of Variance

Var(c) = 0 (constant has zero variance)
Var(aX + b) = a²·Var(X) (adding a constant doesn't change spread)
Var(X + Y) = Var(X) + Var(Y) if X, Y are independent
Var(X) ≥ 0 always

For 2 coins: E(X²) = 0²×(1/4) + 1²×(1/2) + 2²×(1/4) = 0 + 1/2 + 1 = 3/2

Var(X) = 3/2 − 1² = 1/2; σ = 1/√2 ≈ 0.707

Die Example (Verified)

X = face value of a fair die. E(X) = (1+2+3+4+5+6)/6 = 21/6 = 7/2

E(X²) = (1+4+9+16+25+36)/6 = 91/6

Var(X) = 91/6 − (7/2)² = 91/6 − 49/4 = 182/12 − 147/12 = 35/12

5. Finding k in a Probability Distribution

A common problem type: given P(X=xᵢ) = cᵢk, find k using Σpᵢ = 1.

Example: P(X=1) = k, P(X=2) = 2k, P(X=3) = 3k

k + 2k + 3k = 1 → 6k = 1 → k = 1/6

E(X) = 1·(1/6) + 2·(2/6) + 3·(3/6) = 1/6 + 4/6 + 9/6 = 14/6 = 7/3

E(X²) = 1·(1/6) + 4·(2/6) + 9·(3/6) = 1/6 + 8/6 + 27/6 = 36/6 = 6

Var(X) = 6 − (7/3)² = 6 − 49/9 = 54/9 − 49/9 = 5/9

Mean & Variance — Computation Checklist

Step	Action	Formula
1	Verify distribution	Σpᵢ = 1, pᵢ ≥ 0
2	Compute E(X)	Σxᵢpᵢ
3	Compute E(X²)	Σxᵢ²pᵢ
4	Compute Var(X)	E(X²) − [E(X)]²
5	Standard deviation	σ = √Var(X)

JEE/Board Power Tips

Var(X) = E(X²) − [E(X)]² — never compute Var(X) = Σ(xᵢ − μ)²pᵢ directly in exams. The formula E(X²) − [E(X)]² is always faster and less error-prone.
Var(aX + b) = a²·Var(X), NOT a²·Var(X) + b. Adding a constant shifts the distribution but does NOT change the spread — variance is unchanged by adding b.
E(X + Y) = E(X) + E(Y) always — even for dependent RVs. Linearity of expectation holds universally. But Var(X+Y) = Var(X) + Var(Y) only if X, Y are independent.
When finding k: always use Σpᵢ = 1, never try to guess. Then verify each pᵢ ≥ 0 to confirm it's a valid distribution.

Practice Questions

Q1 (Board): A random variable $X$ has the following probability distribution:

$X : 0, 1, 2, 3$
$P (X) : k, 2 k, 3 k, 4 k$

Find the value of $k$ . Hence, find the expected value $E (X)$ and the variance $V a r (X)$ .

Explanation:

Step 1: Find $k$
The sum of all probabilities in a distribution must equal 1 ( $Σ P (x_{i}) = 1$ ):
$k + 2 k + 3 k + 4 k = 1$
$10 k = 1 ⟹ k = \frac{1}{10}$

Step 2: Find Expected Value $E (X)$
$E (X) = Σ [x_{i} \cdot P (x_{i})]$
$E (X) = 0 (\frac{1}{10}) + 1 (\frac{2}{10}) + 2 (\frac{3}{10}) + 3 (\frac{4}{10})$
$E (X) = 0 + \frac{2}{10} + \frac{6}{10} + \frac{12}{10} = \frac{20}{10} = 2$

Step 3: Find Variance $V a r (X)$
First, find $E (X^{2}) = Σ [x_{i}^{2} \cdot P (x_{i})]$ :
$E (X^{2}) = 0^{2} (\frac{1}{10}) + 1^{2} (\frac{2}{10}) + 2^{2} (\frac{3}{10}) + 3^{2} (\frac{4}{10})$
$E (X^{2}) = 0 + \frac{2}{10} + \frac{12}{10} + \frac{36}{10} = \frac{50}{10} = 5$

$V a r (X) = E (X^{2}) - [E (X)]^{2}$
$V a r (X) = 5 - (2)^{2} = 5 - 4 = 1$

Q2 (JEE Main level): If $X$ is a random variable with $E (X) = 3$ and $E (X^{2}) = 13$ , find $V a r (X)$ and $V a r (2 X - 5)$ .

Explanation:

Using the standard variance formula:
$V a r (X) = E (X^{2}) - [E (X)]^{2}$
$V a r (X) = 13 - (3)^{2} = 13 - 9 = 4$

Using the linear transformation property of variance, $V a r (a X + b) = a^{2} \cdot V a r (X)$ :
$V a r (2 X - 5) = 2^{2} \cdot V a r (X)$
$V a r (2 X - 5) = 4 \times 4 = 16$

(Note: Adding or subtracting a constant like $- 5$ simply shifts the distribution but does not affect its spread/variance. Multiplying by a constant $a$ scales the variance by $a^{2}$ ).

Q3 (Board): Construct the probability distribution of $X$ , the number of tails, when a fair coin is tossed 3 times. Find the expected value $E (X)$ and the variance $V a r (X)$ .

Explanation:

When tossing 3 coins, the total number of outcomes in the sample space $n (S) = 2^{3} = 8$ .
The random variable $X$ (number of tails) can take values $0, 1, 2, or 3$ .

Probability Distribution:
$P (X = 0) = \frac{1}{8}$ (HHH)
$P (X = 1) = \frac{3}{8}$ (HHT, HTH, THH)
$P (X = 2) = \frac{3}{8}$ (HTT, THT, TTH)
$P (X = 3) = \frac{1}{8}$ (TTT)
(Check: Sum of probabilities = 1 ✓)

Expected Value $E (X)$ :
$E (X) = 0 (\frac{1}{8}) + 1 (\frac{3}{8}) + 2 (\frac{3}{8}) + 3 (\frac{1}{8})$
$E (X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2}$

Variance $V a r (X)$ :
$E (X^{2}) = 0^{2} (\frac{1}{8}) + 1^{2} (\frac{3}{8}) + 2^{2} (\frac{3}{8}) + 3^{2} (\frac{1}{8})$
$E (X^{2}) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$

$V a r (X) = E (X^{2}) - [E (X)]^{2}$
$V a r (X) = 3 - {(\frac{3}{2})}^{2} = 3 - \frac{9}{4} = \frac{12}{4} - \frac{9}{4} = \frac{3}{4}$

(Optional: Standard Deviation $σ = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \approx 0.866$ )

Q4 (MCQ): A random variable $X$ has $E (X) = 5$ and $E (X^{2}) = 30$ . The standard deviation of $X$ is:

A) $\sqrt{5}$
B) $5$
C) $\sqrt{30}$
D) $25$

Answer: A) $\sqrt{5}$

Explanation:
First, find the variance using the formula: $V a r (X) = E (X^{2}) - [E (X)]^{2}$
$V a r (X) = 30 - (5)^{2} = 30 - 25 = 5$
Standard Deviation ( $σ$ ) is the positive square root of the variance:
$σ = \sqrt{V a r (X)} = \sqrt{5}$

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CategoryProbability

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Random Variables & Probability Distributions

1. Random Variable — Definition and Types

2. Probability Distribution (PMF)

Conditions for a Valid Probability Distribution

Example — Tossing 2 Coins, X = Number of Heads

Cumulative Distribution Function (CDF)

3. Mean (Expected Value) of a Random Variable

Properties of Expectation

4. Variance and Standard Deviation

Properties of Variance

Die Example (Verified)

5. Finding k in a Probability Distribution

Mean & Variance — Computation Checklist

Practice Questions

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