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Using the assumed values from the classroom example where $A$ denotes alcohol dependence and $B$ denotes being a man, you divide the probability of the intersection $P(A\cap B)$ by the probability of the conditioning event $P(B)$. With $P(A\cap B)=0.0225$ and $P(B)=0.5$, the calculation is $0.0225/0.5=0.045$, or $4.5\%$.

Conditions: $P(A\cap B) = 0.0225$.; $P(B) = 0.5$.; These are illustrative inputs, not verified population statistics.

Using the assumed values from the classroom example where $A$ denotes alcohol dependence and $B$ denotes being a man, you divide the probability of the intersection $P(A\cap B)$ by the probability of the conditioning event $P(B)$. With $P(A\cap B)=0.0225$ and $P(B)=0.5$, the calculation is $0.0225/0.5=0.045$, or $4.5\%$.

Conditions: $P(A\cap B) = 0.0225$.; $P(B) = 0.5$.; These are illustrative inputs, not verified population statistics.