Conditional probabilityAQA A-Level Maths: Revision notes

The probability of $A$ given that $B$ has occurred is written $P(A|B)$. The sample space is reduced to $B$, so $P(A|B)=\frac{P(A\cap B)}{P(B)}.$ Rearranging gives the multiplication rule $P(A\cap B)=P(B)\times P(A|B)$. Note that $P(A|B)$ and $P(B|A)$ are usually different: the probability that a Biology student also takes Chemistry is not the probability that a Chemistry student also takes Biology. Always identify the condition (after 'given that') and divide by its probability.

A Venn diagram or two-way table makes conditional probabilities a matter of counting: the denominator is the total in the condition's region or row. Example: of 120 students, 70 study Biology, 50 study Chemistry, 20 both and 20 neither. $P(\text{Chemistry}|\text{Biology})=\frac{20}{70}=\frac27$ and $P(\text{Biology}|\text{not Chemistry})=\frac{50}{70}=\frac57$. When only some totals are given, work out the numbers in each region first (for example 'bicycle only' $=45-12$), and check that they add up to the total.

On a tree diagram the probabilities on the second set of branches are conditional on the first branch. Multiply along branches to find the probability of a combination, and add the results for different routes to the same event. Example: a bag has 5 red and 3 blue counters; two are taken without replacement. $P(\text{blue second}|\text{blue first})=\frac27$. Different colours: $\frac58\times\frac37+\frac38\times\frac57=\frac{15}{28}$. Without replacement, the second-stage probabilities change; with replacement they stay the same, and the events are independent.

Often the tree gives $P(\text{pass}|\text{revised})$ but you want $P(\text{revised}|\text{pass})$. Use the formula: find the intersection from the tree, then divide by the probability of the new condition, found by adding all routes to it. Worked example. $70\%$ of students revise; pass rates are $0.9$ (revised) and $0.4$ (not revised). $P(\text{pass})=0.63+0.12=0.75$. Then $P(\text{revised}|\text{pass})=\frac{0.63}{0.75}=0.84$ and $P(\text{not revised}|\text{fail})=\frac{0.3\times0.6}{0.25}=0.72$.

Events $A$ and $B$ are independent if knowing $B$ does not change the probability of $A$: $P(A|B)=P(A)\quad\text{equivalently}\quad P(A\cap B)=P(A)P(B).$ To test independence, compare $P(A\cap B)$ with $P(A)P(B)$, or $P(A|B)$ with $P(A)$. In the student example, $P(B)P(C)=\frac{35}{144}$ but $P(B\cap C)=\frac{24}{144}$, so the events are not independent. For exactly two passes out of three independent students with pass probability $0.75$: $3\times0.75^2\times0.25=\frac{27}{64}$. Given at least one pass, the probability becomes $\frac{27/64}{63/64}=\frac37$.