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Students often treat "independent" and "mutually exclusive" as two names for "unrelated". They are not. One is about whether events can happen together. The other is about whether one changes the chance of the other. Mix them up and you add when you should multiply, or the reverse. This post gives the two tests, worked examples with dice, cards and coins, and a short routine.
Where is this in the syllabus?
Mutually exclusive events are defined in the NCERT Class 11 chapter Probability, section 14.1.4. Independent events are in section 13.4 of the NCERT Class 12 chapter Probability (2026-27 reprint). The JEE (Main) 2026 syllabus lists "addition and multiplication theorems of probability" in Unit 13. Check the syllabus for your exam year.
What are the two definitions?
Mutually exclusive: A and B have no outcome in common, so A ∩ B = φ and P(A ∩ B) = 0. This is a statement about sets.
Independent: P(A ∩ B) = P(A) × P(B). Equivalently, P(A | B) = P(A): knowing B happened does not change the chance of A. This is a statement about probabilities.
The NCERT Class 12 chapter makes the point directly: "mutually exclusive events never have an outcome common, but independent events, may have common outcome." It adds that two independent events with nonzero probabilities cannot be mutually exclusive, and the other way round.
Why can't they be both?
Suppose P(A) and P(B) are both above zero. If A and B are mutually exclusive, P(A ∩ B) = 0. But P(A) × P(B) is positive. So the independence test fails. Mutually exclusive events are in fact strongly dependent: if one happens, the other's chance drops to zero.
Worked example 1: independent but not exclusive
Roll a fair die. Let E = even number = {2, 4, 6} and F = multiple of 3 = {3, 6}.
P(E) = 1/2, P(F) = 1/3, and E ∩ F = {6}, so P(E ∩ F) = 1/6.
P(E) × P(F) = 1/6. Equal, so independent.
They share the outcome 6, so they are not mutually exclusive.
Worked example 2: exclusive but not independent
Same die. Let G = odd number = {1, 3, 5}. E ∩ G is empty, so E and G are mutually exclusive. P(E ∩ G) = 0, but P(E) × P(G) = 1/4. So they are not independent.
Worked example 3: cards without replacement
Draw two cards one after the other from a 52-card pack, without putting the first back. Let A = first card is an ace and B = second card is an ace.
P(A) = 4/52 = 1/13. By symmetry, P(B) = 1/13 as well.
P(A ∩ B) = (4/52) × (3/51) = 12/2652 = 1/221.
P(A) × P(B) = 1/169, which is not 1/221. So A and B are dependent: the first draw changes what is left.
Worked example 4: adding versus multiplying
P(A) = 0.3 and P(B) = 0.4. Find P(A ∪ B) in two cases.
If A and B are independent: P(A ∩ B) = 0.3 × 0.4 = 0.12, so P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58. Quicker: 1 − P(neither) = 1 − 0.7 × 0.6 = 0.58.
If A and B are mutually exclusive: P(A ∪ B) = 0.3 + 0.4 = 0.7.
Same two numbers, different answers. The question's wording decides which case you are in. For the chance of exactly one under independence: 0.3 × 0.6 + 0.7 × 0.4 = 0.18 + 0.28 = 0.46.
Are pairwise independent events always independent together?
No. NCERT's remark (iv) says three events are mutually independent only if all three pairs multiply and P(A ∩ B ∩ C) = P(A) P(B) P(C).
Toss two fair coins. Let A = first is heads, B = second is heads, C = both show the same face. Each has probability 1/2, and each pair meets in exactly one of the four outcomes, so every pair multiplies to 1/4. But A ∩ B ∩ C is just HH, with probability 1/4, not 1/8. Pairwise independent, but not mutually independent.
Which mistakes cost marks?
Assuming independence from the story. Draws without replacement are dependent.
Adding probabilities for independent events without subtracting P(A ∩ B).
Multiplying probabilities for exclusive events, which gives a positive number for an impossible overlap.
Checking only pairs when a question involves three events.
When you are not sure which rule a problem needs, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. Our post on conditional probability and Bayes' theorem builds directly on this one, and repeated trials and who wins first puts independence to work.
A practice routine
Take ten pairs of events from your textbook exercises and label each pair: exclusive, independent, neither.
Prove each label with a calculation, not a sentence.
Do two three-event questions and test all four conditions.
Note every wrong label and what in the wording misled you.
Start with one die: list every pair of events you can make from "even", "odd", "prime" and "multiple of 3", and test each pair both ways.
