PYQ Vault

JEE Mains Maths · Probability

Random Variables: Mean and Variance

Working with a probability distribution given as a table or a formula: finding an unknown constant, probabilities of ranges, the mean E(X) and the variance E(X²) − (E X)², including the number of special items in a sample drawn without replacement.

All Probability notesJEE Mains Maths strategy

Why this matters

Twenty-three PYQs, sixteen of them multiple choice. Sixteen give or build a distribution and ask for a constant, a probability, the mean or the variance; seven count defective or coloured items in a sample drawn without replacement, where a ready formula saves the table. Two ideas cover the page.

Concept 1 of 2: Distributions, mean and variance

The probabilities add to 1, which fixes any unknown constant. The mean is ∑x P(x)\sum x\,P(x) and the variance is E(X2)−(EX)2E(X^2)-(E X)^2. For an infinite distribution like k(x+1)3−xk(x+1)3^{-x}, the total is an arithmetico-geometric series. For a variable built from other outcomes (a difference of dice, a count of patterns), list its values and their probabilities first.

Definition

  • ∑P(x)=1\sum P(x)=1 fixes the constant.
  • E(X)=∑x P(x)E(X)=\sum x\,P(x); E(X2)=∑x2P(x)E(X^2)=\sum x^2P(x).
  • Var⁡(X)=E(X2)−(EX)2\operatorname{Var}(X)=E(X^2)-(E X)^2, so σ2+μ2=E(X2)\sigma^2+\mu^2=E(X^2).
  • Independent X,YX,Y: Var⁡(X±Y)=Var⁡X+Var⁡Y\operatorname{Var}(X\pm Y)=\operatorname{Var}X+\operatorname{Var}Y.
  • A fair die: mean 72\frac72, variance 3512\frac{35}{12}.

Variance

Var⁡(X)=E(X2)−(E(X))2\operatorname{Var}(X)=E(X^2)-\big(E(X)\big)^2

Worked example

XX takes 0, 1, 2 with probabilities k,2k,3kk,2k,3k. Find E(X)E(X) and Var⁡(X)\operatorname{Var}(X).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q165Moderate

Example 1 · Probability · Random Variables: Mean and Variance

If the mean of the following probability distribution of a random variable XX;
XX02468
P(X)P(X)aa2a2aa+ba+b2b2b3b3b
is 469\frac{46}{9}, then the variance of the distribution is

Square the values, not the probabilities

E(X2)=∑x2P(x)E(X^2)=\sum x^2P(x). Squaring P(x)P(x), or using (∑xP)2(\sum xP)^2 in place of E(X2)E(X^2), gives a wrong variance.
Drill 15 more on distributions, mean and variance

Concept 2 of 2: Special items in a sample without replacement

Draw nn items from NN, of which KK are special, without replacement; XX counts the special ones drawn. Then P(X=k)=(Kk)(N−Kn−k)(Nn)P(X=k)=\frac{\binom Kk\binom{N-K}{n-k}}{\binom Nn}, the mean is nKNn\frac KN, and the variance is nKN⋅N−KN⋅N−nN−1n\frac KN\cdot\frac{N-K}{N}\cdot\frac{N-n}{N-1} — the binomial variance times a correction for not replacing.

Definition

  • P(X=k)=(Kk)(N−Kn−k)(Nn)P(X=k)=\frac{\binom Kk\binom{N-K}{n-k}}{\binom Nn}.
  • E(X)=nKNE(X)=n\frac KN.
  • Var⁡(X)=nKN(1−KN)N−nN−1\operatorname{Var}(X)=n\frac KN\left(1-\frac KN\right)\frac{N-n}{N-1}.
  • With replacement the draws are binomial: drop the last factor.

Sampling without replacement

E(X)=nKN,Var⁡(X)=nKN⋅N−KN⋅N−nN−1E(X)=\frac{nK}{N},\qquad\operatorname{Var}(X)=\frac{nK}{N}\cdot\frac{N-K}{N}\cdot\frac{N-n}{N-1}

Worked example

2 items are drawn without replacement from 8, of which 2 are faulty. Find the mean and variance of the number of faulty items drawn.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 July 2022 · Q179Moderate

Example 2 · Probability · Random Variables: Mean and Variance

A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let XX be the number of white balls, among the drawn balls. If σ2\sigma^{2} is the variance of XX, then 100σ2100\sigma^{2} is equal to

Without replacement changes the variance

The mean is the same with or without replacement, but the variance is smaller without: it carries the factor N−nN−1\frac{N-n}{N-1}.
Drill 6 more on special items in a sample without replacement

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Distributions, mean and variance

    Variance

    Var⁡(X)=E(X2)−(E(X))2\operatorname{Var}(X)=E(X^2)-\big(E(X)\big)^2
  • Special items in a sample without replacement

    Sampling without replacement

    E(X)=nKN,Var⁡(X)=nKN⋅N−KN⋅N−nN−1E(X)=\frac{nK}{N},\qquad\operatorname{Var}(X)=\frac{nK}{N}\cdot\frac{N-K}{N}\cdot\frac{N-n}{N-1}

Watch out for (2)

Test yourself on Probability

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Probability chapter testDrill all 23 questions