Example 1 — P(A | B) vs P(B | A): same overlap, different denominators
StandardProblem
In a class, 12 students play a sport, 8 play music, and 5 do both. (a) Given that a student plays music, what is the probability they also play a sport? (b) Given that a student plays a sport, what is the probability they also play music?
Solution
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For (a) the condition is “plays music,” so the denominator is the 8 music students — not the whole class.
“Given B” shrinks the sample space to B. Here B = music, so only the 8 music players are still in the running.
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Count how many of those 8 also play a sport — that is the 5 who do both. So .
Conditional probability is the overlap divided by the condition’s total: .
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For (b) the condition is now “plays a sport,” so the denominator is the 12 sport students. The overlap is still 5, so .
The shared group (the 5 who do both) did not change — but the denominator did, because we conditioned on a different event.
Answer
and . Same numerator (the 5 who do both), different denominators (8 vs 12) — so the two conditional probabilities are NOT equal.
Takeaway: and share the overlap but divide by different totals. Swapping which event is “given” changes the denominator, so it usually changes the answer.