Statistics and Probability – JEE Main Maths Formula Sheet & Class 11/12 Notes | Mean, Variance, Standard Deviation & All Probability Formulas

Mean — All 6 Methods and Properties

Why All Methods of Computing Mean Are Direct Statistics and Probability JEE Main Formulas

Mean — Unclassified (Individual Series) Data (from PDF — Statistics and Probability):

For n observations x₁, x₂, …, xₙ:

x̄ = (x₁+x₂+…+xₙ)/n = Σxᵢ/n

Mean — Grouped (Discrete) Data (from PDF — Statistics and Probability):

For observations x₁, x₂, …, xₙ with frequencies f₁, f₂, …, fₙ:

x̄ = Σfᵢxᵢ / Σfᵢ = (f₁x₁+f₂x₂+…+fₙxₙ) / (f₁+f₂+…+fₙ)

Shortcut Method (from PDF — Statistics and Probability JEE Main):

Let A = assumed mean (usually the middle term). Define deviation dᵢ = xᵢ – A for each observation.

x̄ = A + Σfᵢdᵢ / Σfᵢ

This reduces computation when observations are large numbers — subtract A first, work with smaller dᵢ values.

Step Deviation Method (from PDF — Statistics and Probability JEE Main):

When data is grouped in equal class intervals of width h, define uᵢ = (xᵢ – A)/h:

x̄ = A + h · (Σfᵢuᵢ / Σfᵢ)

This further simplifies computation when class width h is large — divide dᵢ by h to get even smaller integers uᵢ.

Weighted Mean (from PDF — Statistics and Probability):

For values x₁, x₂, …, xₙ with weights w₁, w₂, …, wₙ:

Weighted Mean = (w₁x₁+w₂x₂+…+wₙxₙ) / (w₁+w₂+…+wₙ) = Σwᵢxᵢ / Σwᵢ

Combined Mean (from PDF — Statistics and Probability JEE Main):

If two data sets have n₁ and n₂ observations with means x̄₁ and x̄₂ respectively:

x̄₁₂ = (n₁x̄₁ + n₂x̄₂) / (n₁ + n₂)

This extends to any number of groups: x̄ = Σnᵢx̄ᵢ / Σnᵢ.

Properties of Mean (from PDF — Statistics and Probability JEE Main):

(1) If each observation is increased (decreased) by a constant k, the new mean = x̄ + k (or x̄ – k).

(2) If each observation is multiplied (divided) by a constant k (k≠0), the new mean = k·x̄ (or x̄/k).

(3) Algebraic sum of deviations from mean is always zero: Σfᵢ(xᵢ – x̄) = 0

(4) Sum of squares of deviations from mean is least (minimum): Σfᵢ(xᵢ – x̄)² is the minimum possible sum of squared deviations from any constant. (Among all measures of central tendency, mean minimises the sum of squared deviations.) Download the Free PDF for Statistics and Probability for all mean formula examples for JEE Main.

Median, Mode, and Empirical Relations

Why Median Formula for Continuous Series and Mode Empirical Relation Are Key Statistics JEE Main Results

Median — Individual Series (from PDF — Statistics and Probability): Arrange in ascending or descending order. Total n observations:

(i) n odd: Median = value of ((n+1)/2)th observation.

(ii) n even: Median = mean of (n/2)th and (n/2+1)th observations.

Median — Discrete Series (from PDF — Statistics and Probability): Prepare cumulative frequency table.

(i) n odd: Median = size of ((n+1)/2)th term in cumulative frequency.

(ii) n even: Median = mean of sizes of (n/2)th and (n/2+1)th terms.

Median — Continuous Series (from PDF — Statistics and Probability JEE Main):

Step I: Prepare cumulative frequency table.

Step II: Locate the median class — the class where (n/2)th observation lies (n = Σfᵢ).

Step III: Apply the median formula:

Median = l + [(n/2 – cf) / f] × h

where: l = lower limit of median class; n = Σfᵢ (total frequency); cf = cumulative frequency of class preceding the median class; f = frequency of the median class; h = class width.

Mode (from PDF — Statistics and Probability):

Mode = value of variate occurring most frequently (having maximum frequency).

Mode of individual series = the value repeated maximum times.

Mode of discrete series = value with highest frequency.

Mode of continuous series (from PDF): Locate modal class = class with maximum frequency. Then:

Mode = l + [(fₘ – fₘ₋₁) / (2fₘ – fₘ₋₁ – fₘ₊₁)] × h

where: l = lower limit of modal class; fₘ = frequency of modal class; fₘ₋₁ = frequency of preceding class; fₘ₊₁ = frequency of succeeding class; h = class width.

If mode falls outside the modal class: Mode = l + [fₘ/(fₘ – fₘ₋₁)] × h (alternative formula).

Empirical Relations (from PDF — Statistics and Probability JEE Main):

Mode = 3 Median – 2 Mean (for moderately skewed distributions)

Rearrangements: Mean = (3 Median – Mode)/2; Median = (Mode + 2 Mean)/3.

Symmetrical distribution: Mean = Median = Mode (all three coincide).

Graphical methods: Median — from ogive; Mode — from histogram; Mean — from frequency curve. Download the Free PDF for Statistics and Probability for all median, mode, and empirical formula examples for JEE Main.

Standard Deviation, Variance, Mean Deviation, and Coefficient of Variation

Why Standard Deviation and Coefficient of Variation Are the Most-Tested Statistics and Probability Topics in JEE Main

Mean Deviation (from PDF — Statistics and Probability):

Mean Deviation about mean: MD(x̄) = Σfᵢ|xᵢ – x̄| / Σfᵢ

Mean Deviation about median: MD(M) = Σfᵢ|xᵢ – M| / Σfᵢ

Coefficient of Mean Deviation = Mean Deviation / Corresponding Average (mean or median)

Relation with SD (from PDF): Mean Deviation = (4/5)σ for moderately non-symmetrical data.

Standard Deviation — Individual (Ungrouped) Data (from PDF — Statistics and Probability JEE Main):

σ = √[Σ(xᵢ – x̄)² / n] (definition form)

Equivalent computation formula: σ = √[(Σxᵢ²/n) – (x̄)²] = √[(Σxᵢ²/n) – (Σxᵢ/n)²]

Standard Deviation — Discrete (Grouped) Data (from PDF — Statistics and Probability):

σ = √[Σfᵢ(xᵢ – x̄)² / Σfᵢ] (definition form)

Computation formula: σ = √[(Σfᵢxᵢ²/Σfᵢ) – (Σfᵢxᵢ/Σfᵢ)²]

Standard Deviation — Grouped Data with Class Marks (from PDF — Statistics and Probability):

Let yᵢ = class mark of each class, ȳ = mean:

σ = √[Σfᵢ(yᵢ – ȳ)² / Σfᵢ]

Variance (from PDF — Statistics and Probability JEE Main): σ² = square of standard deviation. Variance is always ≥ 0. Variance = 0 iff all observations are equal.

Coefficient of Variation (from PDF — Statistics and Probability JEE Main):

CV = (σ / x̄) × 100%

CV is used for comparing variability of two data sets with different units or different means. A series with lesser CV is more consistent (less variable). A series with higher CV is more variable (less consistent).

Standard Normal Distribution Coverage (from PDF — Statistics and Probability):

For a symmetrical distribution:

x̄ ± σ covers 68.27% of observations

x̄ ± 2σ covers 95.45% of observations

x̄ ± 3σ covers 99.73% of observations

Combined Standard Deviation (from PDF — Statistics and Probability JEE Main):

Two data sets: n₁ observations with mean x̄₁ and SD σ₁; n₂ observations with mean x̄₂ and SD σ₂.

Combined mean: x̄₁₂ = (n₁x̄₁ + n₂x̄₂)/(n₁+n₂)

Let d₁ = x̄₁ – x̄₁₂ and d₂ = x̄₂ – x̄₁₂

σ₁₂² = (n₁σ₁² + n₁d₁² + n₂σ₂² + n₂d₂²) / (n₁+n₂) = (n₁(σ₁²+d₁²) + n₂(σ₂²+d₂²)) / (n₁+n₂)

Combined SD: σ₁₂ = √(σ₁₂²). Download the Free PDF for Statistics and Probability for all SD and variance examples for JEE Main.

Probability — Basic Concepts, Event Types, and Classical Probability

Why Probability Definitions, Event Types, and Classical Formula Are the Entry Point for All JEE Main Probability Questions

Random Experiment and Sample Space (from PDF — Statistics and Probability): A random experiment is one whose outcome cannot be predicted with certainty (tossing a coin, throwing a die). The sample space S is the set of all possible outcomes: coin toss S={H,T}; die throw S={1,2,3,4,5,6}. An event is any subset of S.

All Types of Events (from PDF — Statistics and Probability JEE Main):

(1) Simple (Elementary) event: Singleton subset of S — only one outcome.

(2) Compound event: Subset of S with more than one element.

(3) Impossible event: Empty set ∅ — can never occur. P(∅)=0.

(4) Sure (Certain) event: The sample space S itself — always occurs. P(S)=1.

(5) Equally likely events: No one event occurs more often than another.

(6) Mutually exclusive (disjoint) events: Aᵢ∩Aⱼ=∅ for all i≠j — at most one can occur in a trial.

(7) Exhaustive events: A₁∪A₂∪…∪Aₙ=S — at least one must occur in every trial.

Algebra of Events (from PDF — Statistics and Probability):

(i) E' (or Eᶜ or Ē) = complement of E = non-occurrence of E

(ii) E∪F = at least one of E or F occurs

(iii) E∩F = both E and F occur simultaneously

(iv) E'∩F' = neither E nor F occurs

(v) E⊂F = occurrence of E implies F

(vi) E–F = E∩F' = E occurs but not F

(vii) Commutative: E∪F=F∪E; E∩F=F∩E

(viii) De Morgan's: (E∪F)'=E'∩F'; (E∩F)'=E'∪F'

Classical Probability (from PDF — Statistics and Probability JEE Main):

P(E) = n(E)/n(S) = (number of favourable outcomes) / (total number of outcomes)

P(E) = 0 → impossible event. P(E) = 1 → sure event. 0 ≤ P(E) ≤ 1 for all E.

P(∅) = 0; P(S) = 1; P(E) + P(E') = 1 → P(E') = 1 – P(E).

Odds in Favour and Against (from PDF — Statistics and Probability JEE Main):

If event E can happen in a ways and fail in b ways (total a+b equally likely ways):

P(E) = a/(a+b); P(E') = b/(a+b)

Odds in favour of E = a:b = P(E):[1–P(E)]

Odds against E = b:a = [1–P(E)]:P(E)

Conversion: if odds in favour = a:b, then P(E) = a/(a+b). Download the Free PDF for Statistics and Probability for all basic probability examples including cards and dice for JEE Main.

Cards — Key Facts (from PDF — Statistics and Probability): 52 cards total: 4 suits (13 each) — spades (black), clubs (black), hearts (red), diamonds (red). Honour cards: Aces, Kings, Queens, Jacks. Face cards: Kings, Queens, Jacks (12 total). Number cards: 2 through 10 (9 per suit × 4 = 36). Each suit has: 1 Ace + 9 number cards + 3 face cards = 13.

Addition Theorem of Probability — All Forms

Why the Addition Theorem and Its Extended Results Are the Most Directly Tested Statistics and Probability Formulas

Addition Theorem for Mutually Exclusive Events (from PDF — Statistics and Probability):

If E₁, E₂, …, Eₙ are mutually exclusive events:

P(E₁∪E₂∪…∪Eₙ) = P(E₁) + P(E₂) + … + P(Eₙ)

For two mutually exclusive events: P(E∪F) = P(E) + P(F) [since P(E∩F)=0]

Addition Theorem for Any Two Events (from PDF — Statistics and Probability JEE Main):

P(E∪F) = P(E) + P(F) – P(E∩F)

This is the inclusion-exclusion principle for two events.

Addition Theorem for n Events (from PDF — Statistics and Probability JEE Main):

P(E₁∪E₂∪…∪Eₙ) = Σ P(Eᵢ) – Σᵢ<ⱼ P(Eᵢ∩Eⱼ) + Σᵢ<ⱼ<ₖ P(Eᵢ∩Eⱼ∩Eₖ) – … + (–1)ⁿ⁻¹ P(E₁∩E₂∩…∩Eₙ)

Key results derived from Addition Theorem (from PDF — Statistics and Probability JEE Main):

(i) P(exactly one of E, F) = P(E∪F) – P(E∩F) = P(E) + P(F) – 2P(E∩F)

(ii) P(E'∩F') = 1 – P(E∪F) (by De Morgan's); P(E'∪F') = 1 – P(E∩F)

(iii) For n events — Boole's inequality (Union Bound): P(E₁∪E₂∪…∪Eₙ) ≤ P(E₁)+P(E₂)+…+P(Eₙ)

(iv) Bonferroni's inequality: P(E₁∩E₂∩…∩Eₙ) ≥ 1–P(E'₁)–P(E'₂)–…–P(E'ₙ)

(v) P(E₁∩E₂∩…∩Eₙ) ≥ P(E₁)+P(E₂)+…+P(Eₙ) – (n–1)

(vi) If E⊂F: P(E) ≤ P(F) (monotonicity of probability)

Other standard probability results (Statistics and Probability — JEE Main):

P(E∩F') = P(E) – P(E∩F) [E occurs but not F]

P(E'∩F) = P(F) – P(E∩F) [F occurs but not E]

P(E∩F) = P(E) + P(F) – P(E∪F) [rearranging addition theorem]

P(exactly one of E,F) = P(E∩F') + P(E'∩F) = P(E)+P(F)–2P(E∩F). Download the Free PDF for Statistics and Probability for all addition theorem examples for JEE Main.

Conditional Probability, Independent Events, Multiplication Law

Why Conditional Probability and Multiplication Law Are the Most Formula-Dense Statistics and Probability Topics in JEE Main

Conditional Probability (from PDF — Statistics and Probability JEE Main):

The probability of event E given that event F has already occurred:

P(E/F) = P(E∩F) / P(F) [P(F) ≠ 0]

Equivalently: P(E/F) = n(E∩F)/n(F) [ratio of favourable cases in F]

Similarly: P(F/E) = P(E∩F) / P(E) [P(E) ≠ 0]

Properties of Conditional Probability (Statistics and Probability JEE Main):

0 ≤ P(E/F) ≤ 1 for any events E, F with P(F)>0.

P(F/F) = 1; P(∅/F) = 0; P(E'/F) = 1 – P(E/F).

P(E₁∪E₂/F) = P(E₁/F) + P(E₂/F) – P(E₁∩E₂/F).

Multiplication Law of Probability (from PDF — Statistics and Probability):

From conditional probability: P(E∩F) = P(F)·P(E/F) = P(E)·P(F/E)

For n events E₁, E₂, …, Eₙ₊₁ with P(E₁∩E₂∩…∩Eₙ)>0:

P(E₁∩E₂∩…∩Eₙ₊₁) = P(E₁)·P(E₂/E₁)·P(E₃/E₁∩E₂)·…·P(Eₙ₊₁/E₁∩E₂∩…∩Eₙ)

Independent Events (from PDF — Statistics and Probability JEE Main): Events E and F are independent if the occurrence of one does not affect the probability of the other:

P(F/E) = P(F) and P(E/F) = P(E)

E and F independent iff P(E∩F) = P(E)·P(F)

Note from PDF: mutually exclusive events with non-zero probability are NOT independent (if one occurs, the other cannot → occurrence of one affects the other).

Extended independence results (from PDF — Statistics and Probability):

Special cases from PDF: (i) If E₁, E₂, …, Eₙ are independent: P(E₁∩E₂∩…∩Eₙ) = P(E₁)·P(E₂)·…·P(Eₙ).

(ii) P(none of E₁,…,Eₙ occur) = [1–P(E₁)]·[1–P(E₂)]·…·[1–P(Eₙ)]

(iii) If p = probability of success in one trial, probability of success in k consecutive independent trials = pᵏ. Download the Free PDF for Statistics and Probability for all conditional probability and independence examples for JEE Main.

Total Probability Rule and Bayes' Theorem

Why Total Probability and Bayes' Theorem Are the Highest-Difficulty Statistics and Probability Formulas in JEE Main

Total Probability Rule (from PDF — Statistics and Probability JEE Main):

Let {E₁, E₂, …, Eₙ} be a partition of S (mutually exclusive and exhaustive events, P(Eᵢ)>0 for all i). For any event A:

P(A) = Σᵢ₌₁ⁿ P(Eᵢ)·P(A/Eᵢ) = P(E₁)·P(A/E₁) + P(E₂)·P(A/E₂) + … + P(Eₙ)·P(A/Eₙ)

Geometric picture (from PDF): A∩E₁, A∩E₂, …, A∩Eₙ are mutually exclusive and their union = A. So P(A) = Σ P(A∩Eᵢ) = Σ P(Eᵢ)·P(A/Eᵢ).

The two-event special case (from PDF — Facts to Remember): For mutually exclusive and exhaustive E and F, and event G: P(G) = P(E)·P(G/E) + P(F)·P(G/F).

Bayes' Theorem (from PDF — Statistics and Probability JEE Main):

Under the same partition {Eᵢ} of S, for any event A with P(A)>0:

P(Eᵢ/A) = P(Eᵢ)·P(A/Eᵢ) / Σⱼ P(Eⱼ)·P(A/Eⱼ) = P(Eᵢ)·P(A/Eᵢ) / P(A)

For the two-event case (E and F mutually exclusive and exhaustive, G any event) from PDF:

P(E/G) = P(E)·P(G/E) / [P(E)·P(G/E) + P(F)·P(G/F)]

Interpretation of Bayes' Theorem (Statistics and Probability — JEE Main):

P(Eᵢ) = prior probability of cause Eᵢ (before observing A).

P(A/Eᵢ) = likelihood — probability of observed event A given cause Eᵢ.

P(Eᵢ/A) = posterior probability of cause Eᵢ given that A is observed (updated belief after observation).

Standard JEE Main Bayes problems: "A bag is chosen randomly from several bags; a ball is drawn — find the probability it came from bag k given the ball is red." Method: (i) identify causes Eᵢ (bags); (ii) find P(Eᵢ) (probability of choosing each bag); (iii) find P(A/Eᵢ) (probability of red from bag i); (iv) apply Bayes' formula. Download the Free PDF for Statistics and Probability for all Total Probability and Bayes' theorem examples for JEE Main.

Random Variables, Probability Distributions, and Binomial Distribution

Why Binomial Distribution P(r)=ⁿCᵣpʳqⁿ⁻ʳ with Mean=np and Variance=npq Are Must-Know Statistics and Probability Formulas

Random Variable (from PDF — Statistics and Probability): A random variable is a real-valued function defined on the sample space S. It assigns a number to each outcome of a random experiment. Discrete random variable: takes finite or countably infinite values (e.g., number of heads in coin tosses). Continuous random variable: takes any value in an interval.

Probability Distribution (from PDF — Statistics and Probability JEE Main): A table giving all possible values of a discrete random variable X along with their probabilities P(X=xᵢ) is called the probability distribution of X. Requirements: (i) P(X=xᵢ) ≥ 0 for all i; (ii) ΣP(X=xᵢ) = 1 (sum of all probabilities = 1).

Example from PDF — number of heads when two coins tossed:

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

Binomial Distribution (from PDF — Statistics and Probability JEE Main):

If n independent trials are performed with probability p of success in each trial (q = 1–p = probability of failure), then:

P(X=r) = ⁿCᵣ · pʳ · qⁿ⁻ʳ (r = 0, 1, 2, …, n)

The probabilities of 0, 1, 2, …, n successes are the successive terms of the binomial expansion of (q+p)ⁿ.

Key Binomial Distribution results (from PDF — Statistics and Probability JEE Main):

Mean of binomial distribution = np

Variance of binomial distribution = npq

Standard Deviation of binomial distribution = √(npq)

Note: since p+q=1 and p,q>0: npq < np → variance < mean. Coefficient of variation = √(npq)/(np)×100 = √(q/np)×100.

Special probability results using Binomial Distribution (from PDF — Statistics and Probability):

P(at least one success) = 1 – P(0) = 1 – qⁿ

P(at least two successes) = 1 – [P(0)+P(1)] = 1 – qⁿ – nqⁿ⁻¹p

P(exactly r successes in n trials) = ⁿCᵣpʳqⁿ⁻ʳ [direct formula]

P(at most r successes) = Σₖ₌₀ʳ ⁿCₖpᵏqⁿ⁻ᵏ

Multinomial Theorem for Dice (from PDF — Statistics and Probability JEE Main):

For a die with m faces (1, 2, …, m) tossed n times: the probability that the sum of numbers = p is the coefficient of xᵖ in the expansion of [(x+x²+x³+…+xᵐ)/m]ⁿ = [x(1+x+…+xᵐ⁻¹)/m]ⁿ.

This technique replaces direct enumeration in complex dice-sum problems. Download the Free PDF for Statistics and Probability for all binomial distribution and random variable examples for JEE Main.