Conditional probability and independenceEdexcel International A Level Maths: Revision notes

The probability of $B$ given that $A$ has occurred is written $P(B\mid A)$. Knowing that $A$ has happened reduces the sample space to $A$, so $P(B\mid A)=\frac{P(A\cap B)}{P(A)}.$ Example: $P(A)=0.5$, $P(B)=0.4$, $P(A\cap B)=0.2$ give $P(B\mid A)=\frac{0.2}{0.5}=0.4$ and $P(A\mid B)=\frac{0.2}{0.4}=0.5$. In general, $P(B\mid A)\neq P(A\mid B)$. Divide by the probability of the event that is given, which is the one after the vertical line.

Rearranging the definition gives the multiplication rule $P(A\cap B)=P(A)\,P(B\mid A).$ This is used for dependent events, such as drawing without replacement. Along the branches of a tree diagram you multiply, and between alternative routes you add. Example: a bag has 5 red and 3 blue counters, and two are drawn without replacement. $P(\text{both red})=\frac58\times\frac47=\frac{5}{14}$ and $P(\text{different colours})=\frac58\times\frac37+\frac38\times\frac57=\frac{15}{28}$. After the first draw the numbers in the bag change, so the second probability is conditional.

Events $A$ and $B$ are independent if one occurring does not affect the probability of the other. Equivalent conditions are $P(B\mid A)=P(B),\quad P(A\mid B)=P(A),\quad P(A\cap B)=P(A)P(B).$ To test for independence, calculate $P(A)\times P(B)$ and compare it with $P(A\cap B)$. If they are equal the events are independent. Example: $0.5\times0.4=0.2=P(A\cap B)$, so $A$ and $B$ are independent. Then $P(A\mid B')$ is also $0.5$. If $A$ and $B$ are independent then so are $A$ and $B'$, $A'$ and $B$, and $A'$ and $B'$.

These are different ideas and are often confused. Mutually exclusive events cannot occur together: $P(A\cap B)=0$. Independent events satisfy $P(A\cap B)=P(A)P(B)$. If both events have non-zero probability, they cannot be both: exclusive events are strongly dependent, because if one occurs the other certainly does not. For independent events the addition rule becomes $P(A\cup B)=P(A)+P(B)-P(A)P(B)$. Example: $P(A)=0.3$ and $P(A\cup B)=0.58$ give $0.58=0.3+b-0.3b$, so $P(B)=b=0.4$.

Tree diagrams and tables help with context problems. To find a conditional probability, find the joint probability and divide by the probability of the given event. Example: 2% of people have a disease. The test is positive for 95% of those with it and 6% of those without it. $P(D\cap T)=0.02\times0.95=0.019$ and $P(T)=0.019+0.98\times0.06=0.0778$, so $P(D\mid T)=\frac{0.019}{0.0778}=0.244$. Even with an accurate test, a positive result means only about a 24% chance of disease, because the disease is rare and false positives from the large healthy group are numerous.