In short: P(A|B) and P(B|A) are almost never equal — they have different denominators, P(B) versus P(A). Swapping them is the single most common conditional-probability error in JEE Advanced. They are linked only through Bayes' theorem: P(B|A) = P(A|B)·P(B) / P(A), so you can convert one to the other but you can never just transpose the letters.
If you are aiming for IIT, the difference between P(A|B) and P(B|A) is something you must feel in your bones, not just memorise. The bar in "given" tells you what you are conditioning on — the world you have shrunk down to. P(A|B) lives inside the world where B already happened; P(B|A) lives inside the world where A already happened. Those are different worlds with different sizes, so the two numbers are different. Treating them as interchangeable is called conditional probability transposition, and JEE setters design distractors specifically to catch it.
The reliable method
Use this procedure every time a problem says "given", "if", or "knowing that":
- Read the bar right-to-left. P(A|B) means "the probability of A, inside the world where B is true." The event after the bar is your new sample space.
- Identify which event is the condition. Underline the event after the bar. That event's probability becomes your denominator: P(A|B) = P(A∩B) / P(B).
- Check what the question actually asks. Many JEE stems hand you P(A|B) but ask for P(B|A). These are not the same number — you must convert.
- Convert with Bayes when needed. Use P(B|A) = P(A|B)·P(B) / P(A). Find the marginal P(A) using total probability if it is not given.
- Sanity-check the size. A rare cause can produce a frequent symptom, so P(symptom|cause) can be huge while P(cause|symptom) stays small. If your two conditionals come out equal, suspect an error.
A worked example
A factory line produces components. 2% of all components are genuinely defective. A sensor flags a component as "bad" with probability 0.95 when it is truly defective, and flags it as "bad" with probability 0.10 even when it is fine. A component is flagged bad. Find P(defective | flagged bad), and compare it with P(flagged bad | defective).
Let B = "defective" and A = "flagged bad". We are given:
- P(B) = 0.02, so P(not B) = 0.98
- P(A|B) = 0.95
- P(A|not B) = 0.10
First, the marginal P(A) by total probability:
P(A) = P(A|B)·P(B) + P(A|not B)·P(not B) P(A) = (0.95)(0.02) + (0.10)(0.98) = 0.019 + 0.098 = 0.117
Now apply Bayes:
P(B|A) = P(A|B)·P(B) / P(A) = (0.95)(0.02) / 0.117 = 0.019 / 0.117 ≈ 0.162
So P(A|B) = 0.95 but P(B|A) ≈ 0.162. The two conditionals differ by nearly a factor of six. The method holds because each conditional divides by a different denominator — P(B) = 0.02 for one, P(A) = 0.117 for the other — so transposing the letters silently changes what you divide by.
Common mistakes to avoid
- Reading "P(B|A)" as "P(A|B) reversed in value." Fix: rebuild the fraction from scratch — P(B|A) = P(A∩B)/P(A) — and notice the denominator changed.
- Forgetting to compute the marginal P(A). Fix: when the denominator is not handed to you, expand it with total probability across all causes.
- Assuming a strong P(symptom|cause) implies a strong P(cause|symptom). Fix: a rare cause keeps P(cause|symptom) small even when the test is very accurate — this is the base-rate effect.
- Cancelling P(A∩B) carelessly. Fix: P(A∩B) is symmetric (P(A∩B) = P(B∩A)), but the denominators are not, so only the bottom changes when you flip the bar.
- Trusting "equal-looking" answers. Fix: if P(A|B) and P(B|A) come out identical, that only happens when P(A) = P(B); treat it as a red flag, not a result.
Frequently asked questions
Is P(A given B) the same as P(B given A)? No. P(A|B) divides P(A∩B) by P(B), while P(B|A) divides the same intersection by P(A). They are equal only in the special case P(A) = P(B).
How are P(A|B) and P(B|A) related? Through Bayes' theorem: P(B|A) = P(A|B)·P(B) / P(A). Knowing one conditional plus the two marginals lets you compute the other.
Why does swapping them lose marks in JEE Advanced? Setters write distractors that equal the transposed value, so a swap lands you exactly on a wrong option that looks deliberately correct. It signals you did not control the sample space.
When can P(A|B) actually equal P(B|A)? Only when the two events are equally likely, P(A) = P(B). Outside that coincidence, expect them to differ — often by a large factor.
Does P(A∩B) change when I flip the condition? No, intersection is symmetric: P(A∩B) = P(B∩A). Only the denominator changes, which is exactly why the two conditionals differ.
Practise this
Find out if this trips you up — [take the free diagnostic](/diagnostic), then work through the [free Socratic lesson](/lesson/start) to lock in the difference for good.