Sum and product lawsEdexcel International A Level Maths: Revision notes

An event is a set of outcomes. $A\cup B$ means $A$ or $B$ (or both), $A\cap B$ means $A$ and $B$, and $A'$ is the complement (not $A$), with $P(A')=1-P(A)$. The sum law (addition law) is $P(A\cup B)=P(A)+P(B)-P(A\cap B).$ The subtraction removes the overlap, which was counted twice. If $A$ and $B$ are mutually exclusive, they cannot both happen, so $P(A\cap B)=0$ and $P(A\cup B)=P(A)+P(B)$. Example: $P(A)=0.4$, $P(B)=0.5$, $P(A\cap B)=0.15$ gives $P(A\cup B)=0.75$, and $P(\text{neither})=1-0.75=0.25$.

The product law (multiplication law) is $P(A\cap B)=P(A)\times P(B\mid A)$, where $P(B\mid A)$ is the probability of $B$ given that $A$ has happened. Events are independent if one happening does not change the chance of the other. Then $P(B\mid A)=P(B)$ and $P(A\cap B)=P(A)\times P(B).$ To test independence, compare $P(A)P(B)$ with $P(A\cap B)$: if they are not equal, the events are not independent. Example: $0.4\times0.5=0.2$, but $P(A\cap B)=0.15$, so $A$ and $B$ are not independent.

A Venn diagram shows how many outcomes lie in each region. Work from the middle outwards: fill the intersection first, then subtract it from each set total to get the 'only' regions, then find the outside region from the overall total. Example: 60 students, 35 study $M$, 28 study $P$, 15 study both. $M$ only $=35-15=20$, $P$ only $=28-15=13$, neither $=60-(20+15+13)=12$. Then $P(M\cup P)=\frac{48}{60}$ and $P(M\cap P)=\frac{15}{60}$. Since $\frac{35}{60}\times\frac{28}{60}=\frac{49}{180}\neq\frac14$, the events are not independent.

A tree diagram shows a sequence of events. Each branch carries a probability, and the probabilities on branches leaving one point add to 1.

• Multiply along the branches to find the probability of a full path (the product law).
• Add the probabilities of the different paths that give the outcome you want (the sum law for mutually exclusive paths). Example: machine $X$ (60%) is defective with probability $0.05$ and machine $Y$ (40%) with probability $0.02$. $P(\text{defective})=0.6\times0.05+0.4\times0.02=0.038$. For a later stage, the branch probabilities may change depending on what has already happened.

With replacement, each item is returned before the next is taken, so the probabilities do not change and the draws are independent. Without replacement, the item is not returned, so the number of items falls and the second probability depends on the first. Example: 5 red and 3 blue discs, two taken without replacement. $P(\text{both red})=\frac58\times\frac47=\frac{5}{14}$. With replacement it would be $\frac58\times\frac58=\frac{25}{64}$. For 'different colours' add both orders: $2\times\frac58\times\frac37=\frac{15}{28}$. For at least one, it is usually quicker to use the complement: $P(\text{at least one red})=1-P(\text{both blue})=1-\frac{3}{28}=\frac{25}{28}$.