Read the condition first
P(A|B) asks for the probability of A given B. When P(B)>0, it equals P(A∩B)/P(B). The denominator represents the restricted reference group, not automatically the original population.
Penn State provides primary teaching notes on conditional probability and Bayes’ theorem. Understand the reference set before attempting reverse-conditioning formulas.
Build an original count example
Suppose 100 fictional registrants include 40 selecting quantitative workshop A, 30 selecting writing workshop B and 18 selecting both. Then 22 select only A, 12 only B and 48 neither. The four disjoint counts sum to 100.
Of the 30 selecting B, 18 also select A: P(A|B)=18/30=0.6. Of the 40 selecting A, 18 select B: P(B|A)=18/40=0.45. The intersection is unchanged but the denominator differs.
Test independence rather than guessing
Here P(A∩B)=0.18 while P(A)P(B)=0.4×0.3=0.12. These differ, so A and B are not independent under this fictional distribution. Independence does not mean that events cannot happen together.
Equivalently compare P(A|B)=0.6 with P(A)=0.4. The association does not explain participants’ reasons or establish that choosing one workshop causes the other choice.
Use complements to check your work
Self-check: P(not A|B)=12/30=0.4, which complements 0.6. Also P(A∪B)=(40+30−18)/100=0.52, equal to 1−48/100. These checks reveal denominator and double-counting mistakes.
Before calculating, write the total, conditioning group, intersection and target. Explain your method using permitted practice questions and complete assessed work independently.
References
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