Discrete Random VariablesFoundation

Distribution tables → mean & variance — interactive Mathematics simulation for IIT-JEE.

Concept

A discrete random variable attaches a number to each outcome; its probability distribution is the table of values and probabilities (each ≥ 0, summing to 1). Everything flows from two moments: the mean E[X] = Σx·p(x) (the long-run average) and the variance Var(X) = E[X²] − (E[X])², measuring spread. JEE's favourite move: hide an unknown k in the table and force Σp = 1 first.

Key formula

E[X] = \sum x_i p_i; \quad \text{Var}(X) = \sum x_i^2 p_i - \left(\sum x_i p_i\right)^2

Derivation

Var(X) = E[(X − μ)²] expands: E[X² − 2μX + μ²] = E[X²] − 2μ·μ + μ² = E[X²] − μ² — the computational shortcut.

Sum of two dice: X = D₁ + D₂ ⟹ E[X] = 3.5 + 3.5 = 7 by linearity (no table needed!). Variance adds too (independence): 35/12 + 35/12 = 35/6.

Scenarios to explore

Discrete Random Variables — Distribution tables → E[X], variance and the unknown-k trick.

Real-world applications

Expected winnings in games of chance — fair-game analysis.
Insurance premiums: expected claim value.
Decision theory: choosing the option with the best expected payoff.

JEE exam tips

Unknown-k tables: FIRST solve Σp = 1, THEN compute moments.
Linearity: E[aX + b] = aE[X] + b; Var(aX + b) = a²Var(X) — b vanishes from variance.
For max/min of dice, build the CDF first: P(max ≤ m) = (m/6)².

Common mistakes

Var = E[X²] − (E[X])², NOT E[X²] − E[X].
Forgetting to normalise: probabilities must sum to 1 before anything else.
E[X²] ≠ (E[X])² — equality holds only for constants.

Exam traps to avoid

Variance can never be negative — a negative answer means an arithmetic slip.
SD has the same units as X; variance has squared units.