Is P(A given B) the same as P(B given A)?
Short answer: Bayes: P(R|C) = P(C|R)·P(R)/P(C) = 1 × 0.1/0.4 = 0.25.
The trap
The wrong idea
“P(A given B) equals P(B given A).”
It sounds reasonable, and a lot of JEE aspirants carry it into the exam. Here's a one-question check, then the explanation that makes it click.
Quick check
Try it before you peek
P(rain) = 0.1, P(clouds) = 0.4, P(clouds | rain) = 1. P(rain | clouds) is:
- 1
- 0.25
- 0.1
Show the answer
B. 0.25
Bayes: P(R|C) = P(C|R)·P(R)/P(C) = 1 × 0.1/0.4 = 0.25.
The click
In general, no. P(A | B) is the probability of A among the cases where B happens; P(B | A) is the probability of B among the cases where A happens. They are measured against different groups, so they are usually different. Mixing them up is called the confusion of the inverse, and in courtrooms it is known as the prosecutor's fallacy.
They are linked by Bayes' theorem: P(A | B) = P(B | A) × P(A) / P(B). Both share the same numerator but divide by different things: P(A | B) = P(A ∩ B)/P(B) and P(B | A) = P(A ∩ B)/P(A). When P(A ∩ B) > 0, they are equal only if P(A) = P(B).
In the question, P(rain) = 0.1, P(clouds) = 0.4 and P(clouds | rain) = 1 (it never rains without clouds). Then P(rain | clouds) = 1 × 0.1 / 0.4 = 0.25. Picture 100 days: 10 are rainy, and all of those are cloudy; 40 days are cloudy in total. Of the 40 cloudy days only 10 are rainy, and 10/40 = 0.25.
A medical-test example shows why it matters. A test can detect 99% of people who have a disease, so P(positive | disease) = 0.99, yet if the disease is rare and a small fraction of healthy people also test positive, P(disease | positive) can be much lower, because most positive results come from the large healthy group. Always ask: "out of which group?"
Got this one? The Knot Scan finds the misconception you don't know you have, in 8 questions.