Conditional probability and tree and Venn diagramsEdexcel A-Level Maths: Revision notes

$P(A\mid B)$ is the probability that $A$ occurs given that $B$ has occurred. Knowing $B$ happened shrinks the sample space to $B$, so $P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$ Rearranged, this is the multiplication law: $P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A).$ In general $P(A\mid B)\neq P(B\mid A)$; they have different denominators. Events are independent exactly when $P(A\mid B)=P(A)$.

Complement: $P(A')=1-P(A)$. Addition law: $P(A\cup B)=P(A)+P(B)-P(A\cap B),$ since $A\cap B$ is counted in both $P(A)$ and $P(B)$. Often a given $P(A\mid B)$ is used first to find $P(A\cap B)$. Example: $P(A)=0.6$, $P(B)=0.5$, $P(A\mid B)=0.4$ gives $P(A\cap B)=0.2$ and $P(A\cup B)=0.9$. Also $P(B\cap A')=P(B)-P(A\cap B)=0.3$, so $P(B\mid A')=\frac{0.3}{0.4}=0.75$.

Fill in the intersection first, then the 'only' regions by subtraction, then the outside region so everything adds to the total (or to $1$). Then a conditional probability is a ratio of two regions. Example: 50 students, 28 take Mathematics, 20 take Physics, 8 both. Mathematics only $=20$, Physics only $=12$, neither $=10$. $P(\text{Physics}\mid\text{Mathematics})=\frac{8}{28}$: restrict to the 28 who take Mathematics. A two-way table works the same way, with the given condition selecting a row or column.

A tree diagram shows stages. Multiply along a path to get the probability of a combined outcome; add the paths that make up an event. The second-stage branches carry conditional probabilities. Example: $P(\text{rain})=0.3$, late with probability $0.4$ if wet and $0.1$ if dry. $P(\text{late})=0.3\times0.4+0.7\times0.1=0.19$. Branch probabilities from one point must add to 1. To 'reverse' the tree, use the conditional formula: $P(\text{rain}\mid\text{late})=\frac{0.12}{0.19}=\frac{12}{19}$.

A disease affects $2\%$ of people. A test is positive for $95\%$ of those with it and $4\%$ of those without. $P(+)=0.02\times0.95+0.98\times0.04=0.0582$. $P(\text{disease}\mid +)=\frac{0.019}{0.0582}=0.326.$ Most positives are false because the healthy group is very large. A second independent test: $P(\text{disease}\mid ++)=\frac{0.02\times0.95^2}{0.02\times0.95^2+0.98\times0.04^2}=0.920$. Use the full-denominator method every time: sum all routes to the given outcome.