Probability Class 12 Notes: Conditional Probability, Bayes’ Theorem & More

These probability class 12 notes compress Chapter 13 (Probability) from the NCERT Class 12 Mathematics Part II textbook into a revision-ready form. You get conditional probability, the multiplication rule, independent events, the theorem of total probability and Bayes’ theorem — each with definitions, formulas, worked steps and exam pointers.

The chapter builds like a ladder: conditional probability (NCERT, p. 408) → its three properties (p. 408–409) → the multiplication rule (p. 415) → independent events (p. 418) → the partition and theorem of total probability (p. 423–425) → Bayes’ theorem (p. 425–426). The chapter introduction (p. 406) also previews random variables and the binomial distribution, but the exam-critical core is the conditional-probability machinery covered here.

This page ends with a one-page formula sheet, three worked examples with fresh numbers, a common-mistakes table, exam pointers and a five-minute recap. If you need the full set, browse all Class 12 Maths notes or the complete Class 12 notes collection.

The Chapter Map: From Conditional Probability to Bayes’ Theorem

Probability in Class 12 is built in one direction only. Each idea is a step the next idea stands on. Follow this order when you revise:

1. Conditional probability — \( P(E|F) = \frac{P(E \cap F)}{P(F)} \), the probability of E recomputed after F is known (NCERT, p. 408).
2. Three properties — convenient rules for simplifying conditional expressions (p. 408–409).
3. Multiplication rule — the same formula rearranged to find \( P(E \cap F) \) (p. 415).
4. Independent events — the special case \( P(E \cap F) = P(E)P(F) \), where knowing one event changes nothing (p. 418).
5. Partition and the theorem of total probability — splitting the sample space into pieces so any event can be added up (p. 423–425).
6. Bayes’ theorem — reverses the conditional direction, giving the probability of the cause from the effect (p. 425–426).

Everything below follows this exact stack. If you can write Step 1’s formula from memory, Steps 3 and 6 are just rearrangements of it.

Conditional Probability: When the Sample Space Shrinks

Conditional probability answers one question: how does the probability of E change once we already know F happened? The extra information does not change the experiment — it changes the sample space we measure against.

The shrink-the-sample-space analogy. Think of a camera zoom. Before you know F occurred, the lens frames the whole sample space S. The moment you learn F occurred, the camera crops the frame down to F — every outcome outside F is out of the picture. You then measure E only inside the cropped frame.

NCERT’s three-coin experiment makes this concrete (p. 407). Toss three fair coins. Let E = ‘at least two heads’ and F = ‘first coin shows tail’. The sample space has 8 equally likely outcomes:

\[ S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\} \]

Once we know F occurred, the frame crops S to F = {THH, THT, TTH, TTT} — four outcomes. Inside that frame only THH belongs to E. So the conditional probability is:

\[ P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{1}{4} \]

Dividing numerator and denominator by the total number of outcomes gives the working definition (NCERT, p. 408):

Definition 1. For events E and F of the same sample space, the conditional probability of E given F is \[ P(E|F) = \frac{P(E \cap F)}{P(F)} \quad \text{provided } P(F) \neq 0 \]

Why \( P(F) \neq 0 \)? Because the formula divides by \( P(F) \). Division by zero is undefined, so conditional probability simply does not exist when the conditioning event has probability zero. This is why examiners test it — Exercise 13.1 Q16 asks it directly.

Conditional probability also works when outcomes are not equally likely. NCERT Example 7 (p. 412–413) tosses a coin; on head you toss again, on tail you throw a die. The tree diagram (Fig 13.1, 13.2) gives branch probabilities \( \frac{1}{4} \) for (H,H) and (H,T) and \( \frac{1}{12} \) for each (T,i).

With F = ‘at least one tail’ and E = ‘die shows a number greater than 4’:

\[ P(F) = \frac{1}{4} + 6 \times \frac{1}{12} = \frac{3}{4}, \quad P(E \cap F) = 2 \times \frac{1}{12} = \frac{1}{6} \]

\[ P(E|F) = \frac{1/6}{3/4} = \frac{2}{9} \]

The lesson: when outcomes carry different weights, compute \( P(E \cap F) \) and \( P(F) \) by adding branch probabilities, never by counting outcomes.

Three Properties That Make Conditional Probability Behave

These three rules (NCERT, p. 408–409) let you simplify conditional expressions before substituting numbers. Each follows directly from the definition.

• Property 1: \( P(S|F) = P(F|F) = 1 \). Because \( P(S \cap F) = P(F) \), the fraction becomes \( P(F)/P(F) = 1 \). The whole space is certain no matter what we condition on.
• Property 2: \( P((A \cup B)|F) = P(A|F) + P(B|F) – P((A \cap B)|F) \). This mirrors the ordinary addition rule; when A and B are disjoint, the last term is 0 and the expression becomes a clean sum.
• Property 3: \( P(E’|F) = 1 – P(E|F) \). Since \( E \cup E’ = S \), Property 1 plus Property 2 for disjoint events gives the complement rule inside the conditional.

These properties are heavily reused in exam questions that give you \( P(F) \) first and ask you to simplify a compound conditional expression before plugging in numbers. Learn Property 1 as the anchor — the other two build on it.

Multiplication Theorem: From Conditional to Joint Probability

The multiplication rule is the conditional formula flipped around. Start from the definition and rearrange (NCERT, p. 415):

\[ P(E \cap F) = P(E) \cdot P(F|E) = P(F) \cdot P(E|F) \]

Why it holds: the definition says \( P(E|F) = P(E \cap F)/P(F) \). Multiplying both sides by \( P(F) \) recovers the joint probability. Nothing new is being assumed — it is the same information, written differently.

For three events the rule extends naturally (p. 416):

\[ P(E \cap F \cap G) = P(E) \cdot P(F|E) \cdot P(G|EF) \]

Why the probabilities change: after the first draw, the composition of the urn or pack has changed, so each conditional probability is computed on the reduced set. NCERT Example 8 (p. 415–416) draws two balls without replacement from an urn with 10 black and 5 white balls. With E = ‘first ball black’ and F = ‘second ball black’:

\[ P(E) = \frac{10}{15}, \quad P(F|E) = \frac{9}{14} \quad \Rightarrow \quad P(E \cap F) = \frac{10}{15} \times \frac{9}{14} = \frac{3}{7} \]

Notice \( P(F|E) = 9/14 \), not \( 10/15 \): one black ball is gone, so 14 balls remain with 9 black. Example 9 (p. 416–417) chains three draws: \( \frac{4}{52} \times \frac{3}{51} \times \frac{4}{50} = \frac{2}{5525} \) for two kings then an ace. The denominator drops by 1 each draw — that falling denominator is the signature of ‘without replacement’.

Independent Events: When Knowing One Tells You Nothing New

Two events are independent when knowing one occurred gives no information about the other. In the definition’s language (NCERT, p. 418):

Definition 2. E and F are independent if \( P(F|E) = P(F) \) and \( P(E|F) = P(E) \), provided both probabilities are nonzero.

Using the multiplication rule, this collapses to the only test you should ever apply:

Definition 3. E and F are independent if \[ P(E \cap F) = P(E) \cdot P(F) \]

Always test, never assume. Context words like ‘with replacement’ or ‘two separate experiments’ suggest independence, but the product test is the proof. NCERT Example 10 (p. 419) checks a die: E = ‘multiple of 3’ = {3,6}, F = ‘even’ = {2,4,6}, and \( P(E \cap F) = 1/6 = (1/3)(1/2) \), so they are independent.

Independent vs mutually exclusive. This is the chapter’s most confused pair. The NCERT remarks (p. 418–419) are worth memorising as a table:

Three events. A, B, C are mutually independent only if all three pairwise products and the triple product hold (p. 419):

• \( P(A \cap B) = P(A)P(B) \)
• \( P(A \cap C) = P(A)P(C) \)
• \( P(B \cap C) = P(B)P(C) \)
• \( P(A \cap B \cap C) = P(A)P(B)P(C) \)

If any one fails, the events are not mutually independent. Two useful fallout results: if E and F are independent, then E and F′, E′ and F, and E′ and F′ are all independent (Example 13, p. 420); and \( P(\text{at least one of A and B}) = 1 – P(A’)P(B’) \) (Example 14, p. 420).

Partition of a Sample Space and the Theorem of Total Probability

Bayes’ theorem needs a way to slice the sample space into manageable pieces. A set of events \( E_1, E_2, \dots, E_n \) is a partition of S if (NCERT, p. 423–424):

• (a) they are pairwise disjoint: \( E_i \cap E_j = \emptyset \) for \( i \neq j \);
• (b) they are exhaustive: \( E_1 \cup E_2 \cup \dots \cup E_n = S \);
• (c) each has positive probability: \( P(E_i) \gt 0 \) for all i.

Memory device — ‘DEP’. A partition must be Disjoint, Exhaustive, and have Positive probability. Say it as ‘the DEP of a partition’ and you will never list only two conditions. The simplest partition is any event E and its complement E′ — they are disjoint, their union is S, and both can have positive probability.

Theorem of total probability. If \( \{E_1, \dots, E_n\} \) is a partition of S and A is any event, then (p. 424–425):

\[ P(A) = \sum_{j=1}^{n} P(E_j) \, P(A|E_j) \]

Why it works: A can be sliced as \( (A \cap E_1) \cup (A \cap E_2) \cup \dots \cup (A \cap E_n) \). Because the \( E_j \) are disjoint, these slices are disjoint too, so their probabilities add. Each slice is \( P(A \cap E_j) = P(E_j)P(A|E_j) \) by the multiplication rule.

The theorem is just addition rule + multiplication rule glued together.

NCERT Example 15 (p. 425) shows the two-term case. With B = ‘strike’ and A = ‘completed on time’: \( P(B) = 0.65 \), \( P(A|B) = 0.32 \), \( P(A|B’) = 0.80 \). Then \[ P(A) = 0.65 \times 0.32 + 0.35 \times 0.80 = 0.208 + 0.28 = 0.488 \]

The partition here is just {B, B′} — the two-term case is the one you will use most often.

Bayes’ Theorem: The Probability of the Cause

Total probability answers ‘given the cause, what is the chance of the effect?’. Bayes’ theorem asks the reverse: given the effect, what is the chance it came from a particular cause? The NCERT two-bag setup (p. 423) motivates it — a red ball is drawn; which bag did it come from?

Bayes’ theorem. If \( E_1, \dots, E_n \) form a partition of S and \( P(A) \gt 0 \), then for each i (NCERT, p. 425):

\[ P(E_i|A) = \frac{P(E_i) \, P(A|E_i)}{\sum_{j=1}^{n} P(E_j) \, P(A|E_j)} \]

Why it holds: by the definition of conditional probability, \( P(E_i|A) = P(A \cap E_i)/P(A) \). The numerator becomes \( P(E_i)P(A|E_i) \) by the multiplication rule, and the denominator is exactly the theorem of total probability. So Bayes’ theorem is one line of algebra on top of the two previous theorems.

Vocabulary (p. 426): the events \( E_i \) are called the hypotheses; \( P(E_i) \) is the priori probability (known before the experiment); \( P(E_i|A) \) is the a posteriori probability (updated after A is observed). Bayes’ theorem is called the formula for the probability of ’causes’.

The layout examiners reward. In every Bayes’ problem, write the numerator as one branch — cause × effect — and the denominator as the total probability — the sum over ALL branches. NCERT Example 16 (p. 426) uses Bag I (3 red, 4 black) and Bag II (5 red, 6 black), each chosen with probability \( 1/2 \):

\[ P(E_2|A) = \frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7} + \frac{1}{2} \times \frac{5}{11}} = \frac{35}{68} \]

Example 19 (p. 428–429) uses three machines: A, B, C produce 25%, 35%, 40% of output with 5%, 4%, 2% defect rates. Given a defective bolt:

\[ P(B_2|E) = \frac{0.35 \times 0.04}{0.25 \times 0.05 + 0.35 \times 0.04 + 0.40 \times 0.02} = \frac{0.0140}{0.0345} = \frac{28}{69} \]

Real-life application: spam filtering. Bayes’ theorem is what fills your email’s spam folder. Suppose 2% of incoming mail is spam, a filter catches 99% of spam, but also falsely flags 1% of genuine mail. If the filter flags a message, the probability it is truly spam is \[ P(\text{spam}|\text{flagged}) = \frac{0.02 \times 0.99}{0.02 \times 0.99 + 0.98 \times 0.01} \approx 0.669 \]

Only about two-thirds — the small false-positive rate matters because genuine mail is so much commoner than spam. NCERT Example 18 (p. 427–428) makes the same point with an HIV test: a 90%-accurate test still gives only about 0.083 probability that a positive result means the disease, because the disease is rare.

This is the counter-intuitive result that makes Bayes’ theorem worth a full question.

Key Terms at a Glance: Definitions and Examples

Formula Sheet: Every Rule on One Page

Probability has no physical units — every value is a fraction or decimal between 0 and 1. Each rule below carries its condition beside it.

Verify every formula against the official NCERT Class 12 Mathematics Part II textbook (Probability, pages 406–435) if you want the exact statement.

Worked Examples with Fresh Numbers

Worked Example 1: Conditional Probability

Method: definition of conditional probability — shrink the sample space to the given event.

Worked Example 2: Multiplication Rule

Method: without-replacement draws — the second probability is computed on the reduced set.

Worked Example 3: Bayes’ Theorem

Method: identify the partition, write numerator as one branch, denominator as the total probability.

Common Mistakes: Independent vs Mutually Exclusive and Other Traps

Exam Notes: The Steps That Earn the Mark

These patterns are observed from how the chapter’s content is conventionally examined — they are habits that protect marks, not predictions of any paper.

• Always write \( P(F) \neq 0 \) before quoting \( P(E|F) = P(E \cap F)/P(F) \). It is part of the definition (p. 408), and showing it signals you know the condition.
• Show the independence product test in full. State \( P(E) \), \( P(F) \), \( P(E \cap F) \) separately, then compare with \( P(E)P(F) \) — the comparison is the mark.
• In Bayes’ problems, label the hypotheses first. Name the partition events, then write the denominator as the full sum of branches before substituting any number.
• When outcomes are not equally likely (like the toss-then-die tree, p. 412), multiply along the tree branches instead of counting outcomes.
• State the final answer as a simplified fraction. A decimal like 0.488 is fine, but a fraction such as \( 28/69 \) shows the exact value.

Skill map — which exercise practises which skill. If you are short on time, target the exercise that matches your weakest skill:

Related chapters worth revisiting after this one: linear programming notes and vector algebra notes — both reuse the same disciplined, step-by-step layout. For the full subject archive, see the CBSE notes index.

Five-Minute Revision Summary

• Conditional probability shrinks the sample space: \( P(E|F) = P(E \cap F)/P(F) \), with \( P(F) \neq 0 \) (p. 408).
• Multiplication rule reverses it: \( P(E \cap F) = P(E)P(F|E) \) (p. 415).
• Independence is the case \( P(E \cap F) = P(E)P(F) \) — test it, never assume it (p. 418).
• Independent ≠ mutually exclusive. One is about probabilities, the other about subsets (p. 418–419).
• A partition slices S into disjoint, exhaustive, positive-probability pieces — remember ‘DEP’ (p. 423).
• Total probability adds the slices: \( P(A) = \sum P(E_j)P(A|E_j) \) (p. 424).
• Bayes’ theorem reverses the direction to give the probability of the cause (p. 425).

Recall cold, at any hour of the night: ‘at least one’ = \( 1 – P(A’)P(B’) \), and Bayes’ numerator = one branch, denominator = all branches.

FAQs: Conditional Probability, Independence and Bayes’ Theorem

Why is P(A|B) not the same as P(B|A)?

Because the denominators differ. \( P(A|B) = P(A \cap B)/P(B) \) while \( P(B|A) = P(A \cap B)/P(A) \). They are equal only when \( P(A) = P(B) \). The numerators are the same intersection, but the conditioning event — the denominator — changes which probability we normalise by.

When do I use the multiplication rule instead of Bayes’ theorem?

Use the multiplication rule when you want the joint probability \( P(E \cap F) \) — both events happening. Use Bayes’ theorem when you want a conditional probability in the reverse direction: you know the effect A and need the probability of the cause \( E_i \). The clue is the wording ‘given that…’ pointing backwards to the cause.

If two events are mutually exclusive, can they ever be independent?

No, not if both have nonzero probability. Mutual exclusiveness forces \( P(A \cap B) = 0 \), but independence requires \( P(A \cap B) = P(A)P(B) \gt 0 \). The two definitions contradict each other unless one event has probability zero.

Why does P(E|F) require P(F) to be nonzero?

Because the formula divides by \( P(F) \): \( P(E|F) = P(E \cap F)/P(F) \). Division by zero is undefined, so the conditional probability does not exist when the conditioning event cannot occur.

In the family with two children, why is the conditional probability 1/3 and not 1/4?

Because the sample space has four equally likely outcomes: {BB, BG, GB, GG}. Given at least one boy, the frame shrinks to three outcomes {BB, BG, GB}, of which only BB is both boys. So \( P(\text{both boys} \mid \text{at least one boy}) = 1/3 \). The mistake comes from forgetting the two mixed orders BG and GB are distinct.

How do I choose the partition when a Bayes’ theorem question does not name it?

Look for the mutually exclusive, exhaustive causes that could produce the observed effect. In a machine problem, the partition is the machines; in a bag problem, the bags; in a disease test, ‘has disease’ and ‘does not have disease’. Check the three DEP conditions — disjoint, exhaustive, positive probability — and you have your partition.

Reference: NCERT Class 12 Mathematics textbook, chapter Probability.