MathsCBSE Class 12 · Probability

Independent Events: Product Rule, Tests and Applications (Class 12)

Updated 15 October 2026 · Fact-checked

Events E and F are independent when knowing one does not change the chance of the other: P(E|F) = P(E). Equivalently, P(E ∩ F) = P(E) · P(F). To test independence, compute both sides. If E and F are independent, so are E and F′, E′ and F, and E′ and F′.

Each cell's area is the product of its row and column probabilities, so E∩F = P(E)·P(F), and the complement cells follow the same rule.

Illustration drawn for QuestionBuddy.

Understand Independent Events

Two events are independent if the occurrence of one does not affect the probability of the other. Formally (NCERT Section 13.4), events $E$ and $F$ of the same experiment are independent if
$P (E \cap F) = P (E) \cdot P (F)$ .
When $P (F) \neq = 0$ this is the same as $P (E ∣ F) = P (E)$ : being told that F happened gives no information about E.

Testing independence. Compute $P (E)$ , $P (F)$ and $P (E \cap F)$ separately, then compare $P (E \cap F)$ with the product. Equal: independent. Not equal: dependent. Do not decide from intuition; for example, for two tosses of a coin, 'first toss is a head' and 'both tosses show the same face' are independent, which surprises many students.

Independent is not mutually exclusive. Mutually exclusive events cannot happen together: $P (E \cap F) = 0$ . Independent events with non-zero probabilities always have $P (E \cap F) = P (E) P (F) > 0$ , so they can happen together. Hence two events with non-zero probabilities cannot be both independent and mutually exclusive. Mutual exclusivity is about sets; independence is about probabilities.

Complements. If $E$ and $F$ are independent, then so are $E$ and $F^{'}$ , $E^{'}$ and $F$ , and $E^{'}$ and $F^{'}$ . So
$P (E^{'} \cap F^{'}) = (1 - P (E)) (1 - P (F))$ .

Three events. $A$ , $B$ , $C$ are (mutually) independent if every pair satisfies the product rule and $P (A \cap B \cap C) = P (A) P (B) P (C)$ . Pairwise independence alone is not enough.

Workhorse formulas for independent events.

At least one: $P (A \cup B \cup C) = 1 - P (A^{'}) P (B^{'}) P (C^{'})$ .
Exactly one of two: $P (A) P (B^{'}) + P (A^{'}) P (B)$ .
Neither: $P (A^{'}) P (B^{'})$ .

Common contexts are separate people (students solving a problem, shooters, players), separate machines or separate trials of an experiment (tosses with replacement). Draws without replacement are usually not independent.

Independent Events: formulas to remember

Definition: $P (E \cap F) = P (E) \cdot P (F)$; The test for independence.
Conditional form: $P (E ∣ F) = P (E), P (F ∣ E) = P (F)$; When the conditioning event has non-zero probability.
Complements: $P (E^{'} \cap F^{'}) = P (E^{'}) P (F^{'}) = (1 - P (E)) (1 - P (F))$; Complements of independent events are independent.
Union of independent events: $P (E \cup F) = 1 - P (E^{'}) P (F^{'})$; Same as $P (E) + P (F) - P (E) P (F)$ .
Exactly one: $P (E) P (F^{'}) + P (E^{'}) P (F)$; Two disjoint cases added.
Three events: $P (A \cap B \cap C) = P (A) P (B) P (C)$; Needed in addition to the three pairwise conditions.

How to solve Independent Events questions

Two kinds of questions: testing independence, and using it to compute probabilities.

1Testing: list the sample space (or use counts), compute $P (E)$ , $P (F)$ and $P (E \cap F)$ separately, then compare $P (E \cap F)$ with $P (E) P (F)$ and state the conclusion.
2Using: note that the events are independent (stated, or because they come from separate people or trials).
3Write each required event in terms of the basic events and their complements, such as 'exactly one' $= A B^{'} \cup A^{'} B$ .
4Multiply probabilities within each case, then add the disjoint cases.
5For 'at least one', use $1 - P (none)$ ; for unknowns, set up an equation from the given value and solve it.
6Check that every probability you find lies between 0 and 1, and reject roots outside this range.

Independent Events in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Two traffic signals

Suppose on your way to school there are two traffic signals that work independently. The first is green when you arrive with probability 0.5 and the second with probability 0.4.

Find the probability that both are green, and the probability that you have to stop at least once.

Try it first, then show the solutionSolution

Both green: $0.5 \times 0.4 = 0.2$ .
Stop at least once is the complement of 'both green': $1 - 0.2 = 0.8$ .

Answer: 0.2; 0.8

Memory hook: Independent 'and' means multiply.

Why backups work

Imagine your laptop's disk fails in a year with probability 0.05 and your cloud backup is lost with probability 0.01, independently.

Find the probability that you lose your files (both fail).

Try it first, then show the solutionSolution

Both fail: $0.05 \times 0.01 = 0.0005$ .
So the files survive with probability $1 - 0.0005 = 0.9995$ .

Answer: 0.0005 (1 in 2000)

Memory hook: Two independent safety nets multiply the small risks into a tiny one.

Tricks to remember Independent Events

Short memory aids. Say each one aloud once, then use it on a question the same day.

Test = product check
Independent if and only if 'together' equals 'product': $P (E \cap F) = P (E) P (F)$ .
Use it when: Every 'are they independent?' question.
Exclusive = can't meet; independent = don't influence
Mutually exclusive events cannot happen together; independent ones can, they just don't affect each other.
Use it when: Assertion-reason questions on the difference.
At least one = 1 minus none
Multiply the failure probabilities, then subtract from 1.
Use it when: Problem solved, target hit, alarm rings.
Flip all, still free
Complement any of the independent events and they stay independent.
Use it when: Using $P (E^{'}) P (F^{'})$ directly.

Independent Events: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Intersection and union · 2 marks · Very short answerBoard level

A and B are independent events with $P (A) = 0.4$ and $P (B) = 0.25$ . Find $P (A \cap B)$ and $P (A \cup B)$ .

Try it first, then show the solutionSolution

$P (A \cap B) = 0.4 \times 0.25 = 0.1$ . (1 mark)
$P (A \cup B) = 0.4 + 0.25 - 0.1 = 0.55$ . (1 mark)

Answer: $P (A \cap B) = 0.1$ , $P (A \cup B) = 0.55$

Example 2: Testing independence with cards · 3 marks · Short answerBoard level

A card is drawn from a well-shuffled pack of 52. Let E: the card is red, F: the card is a king, G: the card is a heart. Are E and F independent? Are E and G independent?

Try it first, then show the solutionSolution

$P (E) = \frac{1}{2}$ , $P (F) = \frac{4}{52} = \frac{1}{13}$ , $P (E \cap F) = \frac{2}{52} = \frac{1}{26}$ (two red kings). (1 mark)
$P (E) P (F) = \frac{1}{26} = P (E \cap F)$ , so E and F are independent. (1 mark)
$P (G) = \frac{1}{4}$ , $P (E \cap G) = P (G) = \frac{1}{4}$ (every heart is red), but $P (E) P (G) = \frac{1}{8}$ . Not equal, so E and G are not independent. (1 mark)

Answer: E and F are independent; E and G are not.

Example 3: Finding an unknown probability · 3 marks · Short answerHarder board level

A and B are independent events with $P (A) = 0.6$ . The probability that exactly one of them occurs is 0.48. Find $P (B)$ and the probability that neither occurs.

Try it first, then show the solutionSolution

Let $P (B) = q$ . Exactly one: $0.6 (1 - q) + 0.4 q = 0.6 - 0.2 q$ . (1 mark)
$0.6 - 0.2 q = 0.48 \Rightarrow q = 0.6$ . (1 mark)
Neither: $P (A^{'}) P (B^{'}) = 0.4 \times 0.4 = 0.16$ . (1 mark)

Answer: $P (B) = 0.6$ ; P(neither) = 0.16

Example 4: Exactly two of three · 3 marks · Short answerHarder board level

A security system has three sensors that work independently and detect an intruder with probabilities 0.9, 0.8 and 0.7. Find the probability that exactly two sensors detect the intruder, and the probability that at least one does.

Try it first, then show the solutionSolution

Exactly two: $0.9 \times 0.8 \times 0.3 + 0.9 \times 0.2 \times 0.7 + 0.1 \times 0.8 \times 0.7$ . (1 mark)
$= 0.216 + 0.126 + 0.056 = 0.398$ . (1 mark)
At least one: $1 - 0.1 \times 0.2 \times 0.3 = 1 - 0.006 = 0.994$ . (1 mark)

Answer: 0.398; 0.994

Example 5: Taking turns until someone hits · 3 marks · Short answerHOTS / competency

A and B take turns shooting at a target, A first. Each shot of A hits with probability $\frac{1}{3}$ and each shot of B with probability $\frac{1}{2}$ , independently. The first to hit wins. Find the probability that A wins.

Try it first, then show the solutionSolution

A wins on the first shot: $\frac{1}{3}$ . A wins on the second round if both miss once and A then hits: $\frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{3}$ . In general each extra round multiplies by $\frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$ . (1 mark)
$P (A wins) = \frac{1}{3} (1 + \frac{1}{3} + \frac{1}{9} + \dots)$ , an infinite geometric series. (1 mark)
$= \frac{1}{3} \cdot \frac{1}{1 - \frac{1}{3}} = \frac{1}{2}$ . (1 mark)

Answer: $\frac{1}{2}$ (the game is fair, despite A's weaker aim, because A shoots first).

Example 6: Case study: two openers · 4 marks · Case-basedHOTS / competency

Suppose in a match, opener P scores a fifty with probability 0.3 and opener Q with probability 0.4, independently.
(i) Find the probability that both score fifties. (1 mark)
(ii) Find the probability that at least one scores a fifty. (1 mark)
(iii) Find the probability that exactly one scores a fifty. (1 mark)
(iv) Given that at least one scored a fifty, find the probability that both did. (1 mark)

Try it first, then show the solutionSolution

(i) $0.3 \times 0.4 = 0.12$ . (1 mark)
(ii) $1 - 0.7 \times 0.6 = 1 - 0.42 = 0.58$ . (1 mark)
(iii) $0.3 \times 0.6 + 0.7 \times 0.4 = 0.18 + 0.28 = 0.46$ . (1 mark)
(iv) 'Both' is part of 'at least one', so $P = \frac{0.12}{0.58} = \frac{6}{29}$ . (1 mark)

Answer: (i) 0.12 (ii) 0.58 (iii) 0.46 (iv) $\frac{6}{29}$

Where students lose marks in Independent Events

The slips examiners see most often, and what to write instead.

Mistake: Confusing independent with mutually exclusive
Mutually exclusive means $P (E \cap F) = 0$ ; independent means $P (E \cap F) = P (E) P (F)$ .
Wrong working
E and F are independent, so $P (E \cap F) = 0$ and $P (E \cup F) = P (E) + P (F)$ .
Right working
$P (E \cap F) = P (E) P (F)$ and $P (E \cup F) = P (E) + P (F) - P (E) P (F)$ .
Do this: Write the defining equation of each term before using it.
Usually costs: The whole question.
Mistake: Deciding independence by intuition
Independence is a numerical condition, and intuition is often wrong.
Wrong working
'Red' and 'king' are related, so not independent.
Right working
$P (red king) = \frac{1}{26} = \frac{1}{2} \times \frac{1}{13}$ , so independent.
Do this: Always compute both sides and compare.
Usually costs: 1 to 2 marks.
Mistake: Missing cases in 'exactly one' or 'exactly two'
Each choice of who succeeds is a separate case.
Wrong working
Exactly two of three sensors: $0.9 \times 0.8 \times 0.3$ only.
Right working
Add the three cases: $0.216 + 0.126 + 0.056 = 0.398$ .
Do this: List the cases (SSF, SFS, FSS) before multiplying.
Usually costs: 1 to 2 marks.
Mistake: Rejecting a valid root
A quadratic may give two probabilities that both lie in $[0, 1]$ .
Wrong working
$p = \frac{1}{3}$ only.
Right working
$p = \frac{1}{3}$ or $\frac{5}{12}$ , since both keep $2 p \leq 1$ .
Do this: Check each root against all the probability limits.
Usually costs: 1 mark.

Exam tips

Independent-events questions appeared in every board paper from 2023 to 2026, often as a 3-mark question (2023 Q30(b), 2024 Q31, 2026 Q31(b)).
The 2026-27 sample paper has two 1-mark items on independence (Q16 on rain over consecutive days, Q18 on 'exactly one of two events'); expect these forms.
In a 'show that' question, write the three values $P (E)$ , $P (F)$ , $P (E \cap F)$ on separate lines, then the comparison and the conclusion.
State 'since the events are independent' before multiplying; it is a step mark in many schemes.
For 'at least one', the complement route is shorter and less error-prone than adding cases.

Practice questions on Independent Events

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If A and B are independent with $P (A) = 0.5$ and $P (B) = 0.2$ , then $P (A \cup B)$ is:
(a) 0.7
(b) 0.6
(c) 0.1
(d) 0.5
Hint
Subtract $P (A) P (B)$ .
Answer
(b) $0.5 + 0.2 - 0.1 = 0.6$ .
Q2 · 2 marks · Very short answerBoard level
A coin is tossed twice. E: the first toss is a head; F: both tosses show the same face. Are E and F independent?
Hint
$S = {H H, H T, T H, T T}$ .
Answer
$P (E) = \frac{1}{2}$ , $P (F) = \frac{2}{4} = \frac{1}{2}$ , $E \cap F = {H H}$ so $P (E \cap F) = \frac{1}{4} = P (E) P (F)$ . Yes, they are independent.
Q3 · 3 marks · Short answerHarder board level
A die is thrown twice. A: the first throw is odd; B: the sum is 7; C: the sum is 6. Show that A and B are independent but A and C are not.
Hint
$n (B) = 6$ , $n (C) = 5$ .
Answer
$P (A) = \frac{1}{2}$ , $P (B) = \frac{6}{36} = \frac{1}{6}$ ; $A \cap B = {(1, 6), (3, 4), (5, 2)}$ , so $P (A \cap B) = \frac{3}{36} = \frac{1}{12} = P (A) P (B)$ : independent. $P (C) = \frac{5}{36}$ ; $A \cap C = {(1, 5), (3, 3), (5, 1)}$ , so $P (A \cap C) = \frac{1}{12}$ , but $P (A) P (C) = \frac{5}{72}$ : not independent.
Q4 · 3 marks · Short answerHarder board level
A and B are independent with $P (A) = 0.3$ and $P (B) = 0.4$ . Find $P (A ∣ B)$ , $P (A \cap B^{'})$ and $P (A^{'} \cup B^{'})$ .
Hint
$A^{'} \cup B^{'}$ is the complement of $A \cap B$ .
Answer
$P (A ∣ B) = P (A) = 0.3$ ; $P (A \cap B^{'}) = 0.3 \times 0.6 = 0.18$ ; $P (A^{'} \cup B^{'}) = 1 - P (A \cap B) = 1 - 0.12 = 0.88$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): An event A with $P (A) = 0$ is independent of every event B.
Reason (R): For any events, $P (A \cap B) \leq P (A)$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What does R give when $P (A) = 0$ ?
Answer
(a). By R, $0 \leq P (A \cap B) \leq 0$ , so $P (A \cap B) = 0 = P (A) P (B)$ ; R is exactly the reason.
Q6 · 2 marks · Very short answerHarder board level
Three archers hit a target independently with probabilities 0.5, 0.6 and 0.8. Find the probability that exactly one of them hits it.
Hint
Add three cases, one for each archer who hits.
Answer
$0.5 \times 0.4 \times 0.2 + 0.5 \times 0.6 \times 0.2 + 0.5 \times 0.4 \times 0.8 = 0.04 + 0.06 + 0.16 = 0.26$ .
Q7 · 3 marks · Short answerHOTS / competency
A and B are independent events with $P (A) = p$ and $P (B) = 2 p$ . If the probability that exactly one of them occurs is $\frac{5}{9}$ , find $p$ .
Hint
Both roots of the quadratic may be valid; check $2 p \leq 1$ .
Answer
$p (1 - 2 p) + 2 p (1 - p) = 3 p - 4 p^{2} = \frac{5}{9}$ , so $36 p^{2} - 27 p + 5 = 0$ , $(3 p - 1) (12 p - 5) = 0$ , giving $p = \frac{1}{3}$ or $p = \frac{5}{12}$ . Both give $2 p \leq 1$ , so both are valid.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Condition for E and F to be independent?Show answer
$P (E \cap F) = P (E) P (F)$ .
2.If E and F are independent, what is $P (E ∣ F)$ ?Show answer
$P (E)$ (when $P (F) \neq = 0$ ).
3.Can two events with non-zero probabilities be both independent and mutually exclusive?Show answer
No; independence would make $P (E \cap F) > 0$ .
4.A and B independent, $P (A) = P (B) = 0.5$ . Find $P (A^{'} \cap B^{'})$ .Show answer
$0.5 \times 0.5 = 0.25$ .
5.What extra condition do three events need for independence?Show answer
$P (A \cap B \cap C) = P (A) P (B) P (C)$ , besides the three pairwise conditions.
6.Are two draws without replacement usually independent?Show answer
No; the first draw changes the contents for the second.

Independent Events: frequently asked questions

What are independent events in probability?

Two events are independent if the occurrence of one does not change the probability of the other. Mathematically, P(E ∩ F) = P(E) × P(F).

What is the difference between independent and mutually exclusive events?

Mutually exclusive events cannot happen together, so P(E ∩ F) = 0. Independent events can happen together, and P(E ∩ F) = P(E) × P(F). With non-zero probabilities, events cannot be both.

How do I prove two events are independent?

Find P(E), P(F) and P(E ∩ F) separately, usually by counting outcomes. If P(E ∩ F) equals P(E) × P(F), the events are independent; otherwise they are not.

Which NCERT section covers independent events?

Section 13.4 (Independent Events) in Chapter 13 of NCERT Mathematics Part II for Class 12, which also covers independence of three events and of complements.

Are complements of independent events independent?

Yes. If E and F are independent, then E and F′, E′ and F, and E′ and F′ are all independent, so you can multiply their probabilities directly.