Probability revision notes

Key Concepts & Definitions

Random Experiment

An experiment whose precise outcome cannot be predicted with certainty in advance, but all possible outcomes are known.

Sample Space (SSS)

The set of all possible outcomes of a random experiment.

Event

Any subset of the sample space.

Axiomatic Probability

Approach formulated by Russian Mathematician A.N. Kolmogorov, treating probability as a function of outcomes of the experiment.

Conditional Probability

The probability of an event EEE occurring given that another event FFF has already occurred.

Partition of a Sample Space

A set of events E1,E2,…,EnE_1, E_2, \dots, E_nE1​,E2​,…,En​ is a partition if they are pairwise disjoint (Ei∩Ej=ϕE_i \cap E_j = \phiEi​∩Ej​=ϕ for i≠ji \neq ji=j), exhaustive (E1∪E2∪⋯∪En=SE_1 \cup E_2 \cup \dots \cup E_n = SE1​∪E2​∪⋯∪En​=S), and have non-zero probabilities (P(Ei)>0P(E_i) > 0P(Ei​)>0).

Priori Probability

The initial, unconditional probability of a hypothesis EiE_iEi​, denoted by P(Ei)P(E_i)P(Ei​).

Posteriori Probability

The conditional probability of a hypothesis EiE_iEi​ calculated after the event AAA has occurred, denoted by P(Ei∣A)P(E_i | A)P(Ei​∣A).

Random Variable

A real-valued function whose domain is the sample space of a random experiment.

Conditional Probability

• Formula: The conditional probability of event $E$ given $F$ has occurred is defined as $P(E|F) = \frac{P(E \cap F)}{P(F)}$, provided $P(F) \neq 0$.
• Meaning: The information that $F$ has occurred reduces the effective sample space from $S$ to $F$. The outcomes favorable to $E$ are strictly the common elements $E \cap F$.
• Properties:
  • Property 1: $P(S|F) = P(F|F) = 1$.
  • Property 2: For any two events A and B, $P((A \cup B)|F) = P(A|F) + P(B|F) - P((A \cap B)|F)$. If A and B are mutually exclusive (disjoint), $P((A \cup B)|F) = P(A|F) + P(B|F)$.
  • Property 3: $P(E'|F) = 1 - P(E|F)$.
  • JEE TIPConditional probability behaves exactly like classical probability over a restricted sample space. All standard probability axioms apply dynamically to conditional scenarios.

Multiplication Theorem on Probability

• Two Events: $P(E \cap F) = P(E) \cdot P(F|E) = P(F) \cdot P(E|F)$, provided $P(E) \neq 0$ and $P(F) \neq 0$.
• Three Events: $P(E \cap F \cap G) = P(E) \cdot P(F|E) \cdot P(G|E \cap F)$.
• Generalization: The rule extends to $n$ events by consecutively chaining conditional probabilities.
• JEE TIPThe multiplication theorem mathematically translates "without replacement" problems. While the chronological order dictates how you build the equation, the final probability value is independent of the order in which the intersections are evaluated.

Independent Events

• Definition: Two events $E$ and $F$ are independent if the probability of occurrence of one is completely unaffected by the occurrence of the other. Mathematically: $P(E|F) = P(E)$ and $P(F|E) = P(F)$.
• Multiplication Rule for Independent Events: $P(E \cap F) = P(E) \cdot P(F)$.
• Mutual Independence of Three Events: Events A, B, and C are mutually independent if and only if ALL the following hold:
  • $P(A \cap B) = P(A)P(B)$
  • $P(B \cap C) = P(B)P(C)$
  • $P(A \cap C) = P(A)P(C)$
  • $P(A \cap B \cap C) = P(A)P(B)P(C)$.
• JEE TIPPairwise independence (satisfying only the first three conditions) does NOT guarantee mutual independence. You must prove the fourth intersection condition to rule out edge-case dependency.
• Theorems on Independence: If $E$ and $F$ are independent events, then the following pairs are also strictly independent:
  • $E$ and $F'$
  • $E'$ and $F$
  • $E'$ and $F'$
• "At Least One" Rule: If $A$ and $B$ are independent, the probability of occurrence of at least one of $A$ and $B$ is easily solved by using complements: $P(A \cup B) = 1 - P(A')P(B')$.

Theorem of Total Probability & Bayes' Theorem

• Theorem of Total Probability: Let $\{E_1, E_2, \dots, E_n\}$ be a partition of a sample space $S$. For any event $A$ associated with $S$: $P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + \dots + P(E_n)P(A|E_n) = \sum_{j=1}^{n} P(E_j)P(A|E_j)$.
• Bayes' Theorem (Probability of Causes): If $E_1, E_2, \dots, E_n$ form a partition of $S$, and $A$ is any event of non-zero probability, the reverse conditional probability of hypothesis $E_i$ given that $A$ has occurred is: $P(E_i|A) = \frac{P(E_i)P(A|E_i)}{\sum_{j=1}^{n} P(E_j)P(A|E_j)}$.
• JEE TIPBayes' Theorem problems can always be identified by their reverse-chronological question structure. Identify the "Partition Events" ($E_i$) as the mutually exclusive "paths/causes" that happened in the past. Identify the "Observed Event" ($A$) as the current final outcome.

Random Variables & Probability Distributions (Including JEE Advanced Additions)

• Random Variable ($X$): A real-valued function $X: S \to \mathbb{R}$. Multiple random variables can be defined on the same sample space.
• Probability Distribution: A mapping listing all possible values of $X$ ($x_1, x_2, \dots, x_n$) alongside their respective probabilities ($p_1, p_2, \dots, p_n$). Note that $\sum p_i = 1$.
• Expectation (Mean): $\mu = E(X) = \sum_{i=1}^{n} x_i p_i$.
• Variance: $Var(X) = \sigma^2 = E(X^2) - [E(X)]^2 = \sum x_i^2 p_i - \mu^2$. Standard Deviation is $\sigma = \sqrt{Var(X)}$.
• Binomial Distribution: For $n$ independent Bernoulli trials (each with identical success probability $p$ and failure probability $q = 1-p$), the probability of exactly $r$ successes is: $P(X=r) = \binom{n}{r} p^r q^{n-r}$.
  • Mean of Binomial Distribution = $np$
  • Variance = $npq$
• JEE TIPTrap 9 - Games of Chance: Questions involving "infinite alternating throws" until a winner emerges frequently resolve into infinite geometric progressions. Remember the sum of an infinite G.P. is $S_{\infty} = \frac{a}{1-r}$ where $|r| < 1$.

Advanced Combinatorial & Geometrical Probability (JEE Advanced Additions)

• Inclusion-Exclusion Principle (Poincaré's Theorem): $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$.
• Derangements: If $n$ distinct objects are intended for $n$ distinct designated spots, the probability that NONE of the objects occupy their correct spots is: $P(\text{Derangement}) = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!}$.
• Geometrical Probability: When the sample space is continuous (represented by a 1D length, 2D area, or 3D volume), discrete counting fails. Instead: $P(E) = \frac{\text{Measure (Length/Area/Volume) of favourable region E}}{\text{Total Measure of the Sample Space}}$
  • JEE TIPIf a question involves selecting a random point inside a circle or interval, translate the algebraic constraints into geometric inequalities and integrate/compute the resulting areas.

Conditions & Limitations

• Conditional Probability Limits: The formula for $P(E|F)$ strictly requires $P(F) \neq 0$. You cannot condition on an impossible event.
• Multiplication Rule Limits: Evaluating sequential conditional probabilities like $P(F|E)$ inherently limits the application to cases where $P(E) \neq 0$.
• Independence Definition: Determining $P(E|F) = P(E)$ is only mathematically defined if the given condition $P(F) \neq 0$.
• Partition Requirement in Bayes' Theorem: The exhaustive set of hypotheses ($E_i$) must be strictly pairwise disjoint. The total probability denominator collapses if the hypotheses overlap ($E_1 \cap E_2 \neq \phi$) or fail to cover the entire possibility space ($E_1 \cup E_2 \dots \cup E_n \neq S$).

Previous Year JEE Topics

• Bayes' Theorem Applications: The absolute highest-yield topic. Disguised as medical testing scenarios (false positives/negatives), factory machine outputs, or drawing balls from dynamically chosen urns.
• Infinite Geometric Progressions: Probability of winning in sequential games where players alternate turns indefinitely until a specific condition is met.
• Inclusion-Exclusion: Complex set-theory scenarios calculating probabilities with 3 or more overlapping intersections.
• Binomial Distribution Parameters: Using relationships between Mean ($np$) and Variance ($npq$) to solve for number of trials or exact success probabilities.
• Geometrical Probability: Selecting random coordinates $(x,y)$ inside a specified geometric bound (like a circle or square) that must satisfy secondary equations.

Top 10 JEE MCQ Traps

Misconception → Correct Understanding

1. Trap 1 - The Pairwise vs. Mutual Trap

• [JEE TIP] Trap 1 - The Pairwise Independence Fallacy:

  • Misconception: If three events $A, B,$ and $C$ are verified to be pairwise independent ($P(A \cap B) = P(A)P(B)$, $P(B \cap C) = P(B)P(C)$, and $P(A \cap C) = P(A)P(C)$), they are automatically mutually independent as a group.
  • Correct Understanding: Pairwise independence is a necessary condition but is completely insufficient on its own. To declare a set of events mutually independent, you must explicitly check the joint condition: $P(A \cap B \cap C) = P(A)P(B)P(C)$. Examiners often construct problems where the pairs are independent but the trio is not.

• [JEE TIP] Trap 2 - Independent vs. Mutually Exclusive Polar Opposites:

  • Misconception: If two events are independent, it means they are disjoint and share no overlapping outcomes in the sample space.
  • Correct Understanding: Independence and mutual exclusivity are completely opposite concepts. If two events have non-zero probabilities, they cannot be both independent and mutually exclusive. Mutual exclusivity means if one happens, the other cannot ($P(A \cap B) = 0$). Independence requires that the occurrence of one does not affect the other, which demands a non-zero intersection ($P(A \cap B) = P(A)P(B) > 0$).

• [JEE TIP] Trap 3 - Without Replacement Fraction Lock:

  • Misconception: Drawing successive items from a container without replacement allows you to treat the probabilities of later draws as static fractions that are independent of earlier outcomes.
  • Correct Understanding: Sampling without replacement introduces dynamic conditional dependence because the sample space shrinks with each step. You cannot simply multiply unconditioned probabilities. You must strictly apply the Multiplication Theorem of Probability: $P(E \cap F) = P(E) \cdot P(F|E)$, where the probability of the second event is recalculated based on what was removed in the first step.

• [JEE TIP] Trap 4 - The Bayes' Denominator Omission:

  • Misconception: When calculating the posterior inverse probability $P(E_i|A)$ using Bayes' Theorem, it is sufficient to evaluate only the product of the numerator terms: $P(E_i) \cdot P(A|E_i)$.
  • Correct Understanding: The numerator product evaluates only the joint intersection probability $P(E_i \cap A)$. To find the true conditional probability, you must normalize the fraction by dividing by the Total Probability of event $A$. This requires summing up the probabilities of all mutually exclusive and exhaustive pathways that can lead to outcome $A$: $\sum P(E_k) \cdot P(A|E_k)$.

• [JEE TIP] Trap 5 - The Clinical Diagnostic Inversion:

  • Misconception: The probability that a diagnostic test returns a positive result given that a patient has a disease ($P(\text{Test}^+|\text{Disease})$) is identical to the probability that a patient actually has the disease given that they test positive ($P(\text{Disease}|\text{Test}^+)$).
  • Correct Understanding: These two conditional probabilities are fundamentally different. $P(\text{Test}^+|\text{Disease})$ reflects the technical accuracy or sensitivity of the test. Conversely, finding $P(\text{Disease}|\text{Test}^+)$ requires applying Bayes' Theorem. This value is heavily influenced by the background rarity of the disease and can drop drastically if the condition is rare in the general population.

• [JEE TIP] Trap 6 - The Additive "At Least One" Trap:

  • Misconception: To calculate the probability of at least one independent event occurring among a set of events, you can simply add their individual probabilities together: $P(A \cup B) = P(A) + P(B)$.
  • Correct Understanding: Summing individual probabilities directly ignores the intersection and risks generating a value greater than $1$, which violates probability axioms. For independent events, the most efficient and accurate approach is to calculate through the complement: $P(\text{at least one}) = 1 - P(A') \cdot P(B')$, where you subtract the probability of complete failure from unity.

• [JEE TIP] Trap 7 - Sample Space Denominator Inertia:

  • Misconception: When evaluating a conditional probability where an event $F$ is stated to be "given", the denominator of your probability fraction remains equal to the total unconstrained sample space $S$.
  • Correct Understanding: The condition "given $F$" instantly collapses and alters your entire probability universe. The outcomes outside of event $F$ become completely irrelevant. The elements (or total probability) of event $F$ must strictly serve as your new denominator, transforming the calculation to: $P(A|F) = \frac{P(A \cap F)}{P(F)}$.

• [JEE TIP] Trap 8 - Dependent Random Variance Expansion:

  • Misconception: The variance of a sum of two random variables is always equal to the sum of their individual variances: $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y)$.
  • Correct Understanding: This simple additive relationship holds true exclusively if the random variables $X$ and $Y$ are strictly independent. If the variables share a statistical dependency, the equation must incorporate an additional joint term: $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X, Y)$. Forgetting the covariance term is a classic source of errors in advanced statistics questions.

• [JEE TIP] Trap 9 - Alternating G.P. First-Term Misalignment:

  • Misconception: When modeling a game where players alternate turns until someone wins, the first term '$a$' of the infinite Geometric Progression (G.P.) is always equal to the standard single-round winning probability for any player.
  • Correct Understanding: The first term '$a$' of the G.P. must strictly represent the probability of that specific player winning on their very first attempt. While this is straightforward for Player 1, the first term for Player 2 must explicitly incorporate the condition that Player 1 fails their opening turn. For example, if the single-round win probability is $p$ and failure is $q$, Player 2's first term is $qp$, not $p$.

• [JEE TIP] Trap 10 - The Base Rate Blindspot:

  • Misconception: If a medical test or screening device is highly accurate (e.g., $99\%$ reliable), a positive result guarantees that the condition is present, regardless of the initial prior probability $P(E_i)$.
  • Correct Understanding: This is the classic Base Rate Fallacy. Even if the conditional accuracy $P(A|E_i)$ is exceptionally high, if the base rate of the event ($P(E_i)$) is extremely small (e.g., $1$ in $10,000$), the vast majority of positive test results will mathematically turn out to be false positives. You must always allow the prior baseline probability to weight and anchor the Bayes' Theorem equation.