In short: You can multiply P(A)·P(B) to get P(A∩B) only when A and B are independent — when one happening does not change the other's probability. For dependent events you must use the general rule P(A∩B) = P(A)·P(B|A). Assuming independence when it does not hold is one of the costliest JEE probability errors.
"When can you multiply probabilities?" is a question that decides a large fraction of JEE Advanced probability marks. The shortcut P(A∩B) = P(A)·P(B) is seductive because it is so quick — but it is only valid for independent events. Drawing without replacement, conditioning on a prior outcome, or any setup where one event shifts the odds of another breaks independence, and the shortcut quietly gives a wrong answer. The safe habit is to ask "does A change B?" before you ever multiply.
The reliable method
Before multiplying two probabilities, run this check:
- Ask the independence question. Does knowing A occurred change the probability of B? If no, they are independent. If yes, they are dependent.
- If independent, multiply directly. P(A∩B) = P(A)·P(B). This is valid exactly when P(B|A) = P(B).
- If dependent, use the chain rule. P(A∩B) = P(A)·P(B|A). Compute the conditional probability of B after A has happened.
- Watch for replacement. "With replacement" usually restores independence; "without replacement" almost always creates dependence.
- Verify rather than assume. Independence is something you justify from the setup, never something you assume because it is convenient.
A worked example
A bag holds 3 red and 2 blue balls. Two balls are drawn one after another without replacement. Find the probability that both are red.
Let A = "first ball is red" and B = "second ball is red". The temptation is to write P(A)·P(A) = (3/5)(3/5) = 9/25, but that assumes the second draw faces the same odds as the first — which is false, because we did not put the first ball back.
After drawing one red, the bag holds 2 red and 2 blue (4 balls total), so:
- P(A) = 3/5
- P(B|A) = 2/4
Apply the chain rule:
P(A∩B) = P(A)·P(B|A) = (3/5)(2/4) = 6/20 = 3/10 = 0.3
So the correct probability that both balls are red is 3/10, not the 9/25 ≈ 0.36 the independence shortcut would give. The method holds because removing a red ball changes what remains: the second draw is conditioned on the first, so P(B|A) ≠ P(B) and only P(A)·P(B|A) counts the outcome correctly.
Common mistakes to avoid
- Multiplying P(A)·P(B) for dependent events. Fix: use P(A)·P(B|A) whenever one event affects the other.
- Treating "without replacement" as independent. Fix: removing an item changes the pool, so later draws are conditional — switch to the chain rule.
- Confusing independent with mutually exclusive. Fix: independent means P(A∩B) = P(A)·P(B); mutually exclusive means P(A∩B) = 0. They are opposite ideas, not the same.
- Assuming independence because events "feel unrelated". Fix: justify independence from the mechanism (replacement, separate experiments), not intuition.
- Reusing P(A) for the second factor. Fix: recompute the conditional after A occurs — the counts have changed.
Frequently asked questions
Can I always multiply P(A) and P(B) to get P(A and B)? No. P(A∩B) = P(A)·P(B) holds only when A and B are independent. For dependent events use P(A∩B) = P(A)·P(B|A).
How do I know if two events are independent? They are independent if P(B|A) = P(B), meaning A occurring does not change B's probability. Equivalently, P(A∩B) = P(A)·P(B).
Are draws without replacement independent? Almost never. Removing an item changes the remaining pool, so each later draw is conditional on the earlier ones — use the chain rule, not simple multiplication.
What is the difference between independent and mutually exclusive? Independent events can occur together and satisfy P(A∩B) = P(A)·P(B). Mutually exclusive events cannot occur together, so P(A∩B) = 0. They describe different relationships.
What rule replaces multiplication when events are dependent? The general multiplication rule: P(A∩B) = P(A)·P(B|A). It always works, and reduces to P(A)·P(B) when the events happen to be independent.
Practise this
Find out if this trips you up — [take the free diagnostic](/diagnostic), then work through the [free Socratic lesson](/lesson/start) to learn exactly when multiplying is allowed.