This sheet covers the Probability Class 12 formulas from NCERT, grouped by topic: conditional probability and its properties, the multiplication rule, independent events, partition of a sample space, the theorem of total probability, and Bayes’ theorem.
Each formula below carries its symbol meanings, the condition under which it is valid, and when to reach for it. Worked examples use original numbers and show the formula being selected before each substitution. For detailed explanations and derivations of these results, see the Probability Class 12 notes.
Formulas at a Glance
| Purpose (what you are finding) | Formula |
|---|---|
| Conditional probability of E given F | \( P(E|F) = \dfrac{P(E \cap F)}{P(F)} \), \( P(F) \neq 0 \) |
| Probability that both E and F occur (multiplication rule) | \( P(E \cap F) = P(E) P(F|E) = P(F) P(E|F) \) |
| Intersection of three events | \( P(E \cap F \cap G) = P(E) P(F|E) P(G|EF) \) |
| E and F independent (definition) | \( P(E \cap F) = P(E) P(F) \) |
| Probability of at least one of two independent events | \( P(A \cup B) = 1 – P(A’) P(B’) \) |
| Total probability with a partition \( E_1, \dots, E_n \) | \( P(A) = \sum_{j=1}^{n} P(E_j) P(A|E_j) \) |
| Bayes’ theorem (probability of a cause) | \( P(E_i|A) = \dfrac{P(E_i) P(A|E_i)}{\sum_{j=1}^{n} P(E_j) P(A|E_j)} \) |
All Formulas, Grouped by Topic
Conditional Probability
If E and F are events of a sample space, the conditional probability of E given that F has already occurred is defined as (NCERT, p. 2):
\[ P(E|F) = \frac{P(E \cap F)}{P(F)}, \quad P(F) \neq 0 \]
For equally likely elementary outcomes, the same rule can be counted directly using numbers of favourable outcomes (NCERT, p. 1):
\[ P(E|F) = \frac{n(E \cap F)}{n(F)} \]
Properties of Conditional Probability
The complement rule (NCERT, p. 3):
\[ P(E’|F) = 1 – P(E|F) \]
The union rule for conditional probabilities (NCERT, p. 3):
\[ P((A \cup B)|F) = P(A|F) + P(B|F) – P((A \cap B)|F) \]
When A and B are disjoint, the last term is zero:
\[ P((A \cup B)|F) = P(A|F) + P(B|F) \]
Multiplication Rule of Probability
Rearranging the conditional probability definition gives the intersection formula (NCERT, p. 10):
\[ P(E \cap F) = P(E) \cdot P(F|E) = P(F) \cdot P(E|F), \quad P(E) \neq 0,\; P(F) \neq 0 \]
For three events (NCERT, p. 10):
\[ P(E \cap F \cap G) = P(E) \cdot P(F|E) \cdot P(G|EF) \]
Independent Events
Two events are independent exactly when the occurrence of one does not change the probability of the other (NCERT, p. 12):
\[ P(E \cap F) = P(E) \cdot P(F) \]
Equivalently, when the conditional probabilities are unchanged (NCERT, p. 12):
\[ P(E|F) = P(E) \quad (P(F) \neq 0); \qquad P(F|E) = P(F) \quad (P(E) \neq 0) \]
If E and F are independent, then E’ and F, E and F’, and E’ and F’ are also independent (NCERT, p. 15). For two independent events, the probability that at least one occurs (NCERT, p. 15):
\[ P(A \cup B) = 1 – P(A’) P(B’) = P(A) + P(B) – P(A) P(B) \]
Three events A, B, C are mutually independent only when all four products hold (NCERT, p. 13):
\[ P(A \cap B) = P(A) P(B), \quad P(B \cap C) = P(B) P(C), \quad P(A \cap C) = P(A) P(C), \quad P(A \cap B \cap C) = P(A) P(B) P(C) \]
Partition of a Sample Space and Theorem of Total Probability
Events \( E_1, E_2, \dots, E_n \) partition the sample space S when they are pairwise disjoint, their union is S, and each has nonzero probability (NCERT, p. 18):
\[ E_i \cap E_j = \phi \; (i \neq j), \qquad E_1 \cup E_2 \cup \dots \cup E_n = S, \qquad P(E_i) \gt 0 \]
For any event A, the theorem of total probability gives (NCERT, p. 19):
\[ P(A) = \sum_{j=1}^{n} P(E_j) \cdot P(A|E_j) \]
For the simple partition \( \{E, E’\} \) this special case is enough (NCERT, p. 20):
\[ P(A) = P(E) P(A|E) + P(E’) P(A|E’) \]
Bayes’ Theorem
If \( E_1, \dots, E_n \) partition S and A is any event of nonzero probability (NCERT, p. 20):
\[ P(E_i|A) = \frac{P(E_i) \cdot P(A|E_i)}{\sum_{j=1}^{n} P(E_j) \cdot P(A|E_j)}, \qquad i = 1, 2, \dots, n \]
The standard two-cause form (NCERT, p. 20):
\[ P(E|A) = \frac{P(E) P(A|E)}{P(E) P(A|E) + P(E’) P(A|E’)} \]
What Each Symbol Means
| Symbol | What it means | Nature / Unit |
|---|---|---|
| \( P(E) \) | Probability that event E occurs | Dimensionless, \( 0 \leq P(E) \leq 1 \) |
| \( S \) | Sample space of the random experiment | Set of all outcomes |
| \( E’, A’, B’ \) | Complement of an event (event does not occur) | Event |
| \( E \cap F \) | Both E and F occur simultaneously (also written EF) | Event |
| \( E \cup F \) | At least one of E or F occurs | Event |
| \( P(E|F) \) | Conditional probability of E given that F has already occurred | Dimensionless, \( 0 \leq P(E|F) \leq 1 \) |
| \( P(E_i) \) | Prior (a priori) probability of the \( i \)th hypothesis \( E_i \) | Dimensionless |
| \( P(A|E_i) \) | Probability of A given that hypothesis \( E_i \) occurs | Dimensionless |
| \( P(E_i|A) \) | Posterior (a posteriori) probability of \( E_i \) given A | Dimensionless |
| \( n(E) \) | Number of elementary outcomes favourable to E | Count |
When to Use Each Formula
- Conditional probability — use whenever the question fixes one event as already true (“given that”, “if it is known that”). It is valid only when \( P(F) \neq 0 \).
- Complement rule \( P(E’|F) = 1 – P(E|F) \) — use for “at least one”, “not”, or “fails” language once a condition F is fixed.
- Multiplication rule — use to find \( P(E \cap F) \) when draws are made without replacement, or when you know \( P(F|E) \) and need the intersection.
- Independence test — check \( P(E \cap F) = P(E) P(F) \). If the word “independent” is given in the problem, you may multiply probabilities directly; you do not need conditional probabilities. Remember: an independent pair with nonzero probabilities can never be mutually exclusive.
- Total probability — use when the sample space splits into cases (success/failure, machine A/B/C, bag I/II) and you need the overall \( P(A) \).
- Bayes’ theorem — use when the problem gives the probabilities of the “causes” and asks for the reverse probability: “given the observed result A, what is the probability it came from \( E_i \)?”
Worked Examples
Example 1: Direct Conditional Probability
A die is rolled once. Let E be “the number shown is even” and F be “the number shown is greater than 4”. Find \( P(E|F) \).
Step 1: Identify the conditioned event.
We need \( P(E|F) \), so \( F = \{5, 6\} \) is the denominator.
Step 2: List \( E \cap F \).
Even numbers greater than 4: only 6, so \( E \cap F = \{6\} \).
\[ P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{6}}{\frac{2}{6}} = \frac{1}{2} \]
Final answer: \( P(E|F) = \dfrac{1}{2} \).
Example 2: Bayes’ Theorem (Reverse Probability)
A bag contains 3 red and 5 blue balls; another bag contains 2 red and 4 blue balls. One bag is chosen at random and a ball drawn is found to be red. Find the probability that it came from the first bag.
Step 1: Set \( E_1 \) = first bag chosen, \( E_2 \) = second bag chosen, A = red ball drawn.
Since a bag is chosen at random, \( P(E_1) = P(E_2) = \tfrac{1}{2} \).
- Step 1: Find the conditional draws: \( P(A|E_1) = \tfrac{3}{8} \), \( P(A|E_2) = \tfrac{2}{6} = \tfrac{1}{3} \).
- Step 2: Apply Bayes’ theorem.
\[ P(E_1|A) = \frac{P(E_1) P(A|E_1)}{P(E_1) P(A|E_1) + P(E_2) P(A|E_2)} \]
\[ P(E_1|A) = \frac{\tfrac{1}{2} \times \tfrac{3}{8}}{\tfrac{1}{2} \times \tfrac{3}{8} + \tfrac{1}{2} \times \tfrac{1}{3}} = \frac{\tfrac{3}{16}}{\tfrac{3}{16} + \tfrac{1}{6}} = \frac{9}{17} \]
Final answer: \( P(E_1|A) = \dfrac{9}{17} \).
Example 3: Alternating Game (Geometric Series Combination)
A and B throw a die alternately till one of them gets a 2 and wins. If A starts first, find each player’s probability of winning.
Step 1: Let \( S \) = success (a 2 shows) and \( F \) = failure.
Then \( P(S) = \tfrac{1}{6} \), \( P(F) = \tfrac{5}{6} \).
Step 2: A wins on his 1st, 3rd, 5th, …
throw; between A’s turns both players must fail.
\[ P(A \text{ wins}) = \frac{1}{6} + \left(\frac{5}{6}\right)^2 \frac{1}{6} + \left(\frac{5}{6}\right)^4 \frac{1}{6} + \dots = \frac{\tfrac{1}{6}}{1 – \tfrac{25}{36}} = \frac{6}{11} \]
Step 3: Since the game must end, \( P(B \text{ wins}) = 1 – P(A \text{ wins}) \).
\[ P(B \text{ wins}) = 1 – \frac{6}{11} = \frac{5}{11} \]
Final answer: A wins with probability \( \dfrac{6}{11} \), B wins with probability \( \dfrac{5}{11} \).
Common Mistakes to Avoid
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Writing \( P(E|F) = \dfrac{P(F|E)}{P(F)} \) or swapping events | \( P(E|F) = \dfrac{P(E \cap F)}{P(F)} \) | The denominator is always the probability of the event that is given, i.e. F. |
| Dividing by \( P(F) = 0 \) | The formula is defined only when \( P(F) \neq 0 \) | If the given event has probability 0, say “not defined” — as in Exercise 13.1, Q16. |
| Using \( P(E \cap F) = P(E) P(F) \) without the events being independent | The product rule holds only for independent events; otherwise use the multiplication rule with a conditional probability | Verify \( P(E|F) = P(E) \) or state independence in the problem. |
| Forgetting the \( P(E_i) P(A|E_i) \) terms in the denominator of Bayes’ theorem | Denominator is total probability: \( \sum_j P(E_j) P(A|E_j) \) | The denominator must equal \( P(A) \) — it should be calculable independently. |
| Using the mutually exclusive test in place of the independence test | Mutually exclusive means \( P(E \cap F) = 0 \); independent means \( P(E \cap F) = P(E) P(F) \) | Two independent events of nonzero probability can not be mutually exclusive. |
Frequently Asked Questions
Why must \( P(F) \neq 0 \) in conditional probability?
The conditional probability is defined as a ratio with \( P(F) \) in the denominator. If \( P(F) = 0 \), the division is undefined and the formula has no meaning.
What is the difference between independent and mutually exclusive events?
Independent events satisfy \( P(E \cap F) = P(E) P(F) \) — one event does not change the other’s probability. Mutually exclusive events satisfy \( E \cap F = \phi \) — they share no outcome, so \( P(E \cap F) = 0 \). Two independent events with nonzero probabilities cannot be mutually exclusive.
When do I use the theorem of total probability versus Bayes’ theorem?
Use the theorem of total probability when you need the overall \( P(A) \) by summing over the partition cases. Use Bayes’ theorem when you already know A occurred and need the reverse probability \( P(E_i|A) \) — the probability that a particular cause \( E_i \) was responsible.
How do I check my Bayes’ theorem answer?
First find the denominator \( \sum_j P(E_j) P(A|E_j) \), which must equal \( P(A) \). Then confirm that the sum of all the posteriors \( \sum_i P(E_i|A) \) equals 1. Both checks catch a missing term in the denominator.
Practise these formulas on the textbook’s own questions — the NCERT solutions for Probability Class 12 walk through the exercise problems step by step. For the broader chapter’s place in the course, see the Class 12 Maths formulas hub.
Reference: NCERT Class 12 Mathematics textbook, chapter Probability (Rationalised NCERT).
Explore Class 12 Maths Formulas
More for this chapter:
- Probability Notes
Related chapters:
Official source: download the NCERT textbook free from ncert.nic.in.