Mutually exclusive and independent eventsAQA A-Level Maths: Revision notes

Two events are mutually exclusive if they cannot happen at the same time, so $P(A\cap B)=0$. The addition rule for mutually exclusive events is $P(A\cup B)=P(A)+P(B).$ For events that are not mutually exclusive the general addition rule is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, which stops the overlap being counted twice. If $P(A)=0.4$ and $P(B)=0.35$ are mutually exclusive, then $P(A\cup B)=0.75$ and the probability of neither is $0.25$.

Two events are independent if the occurrence of one does not affect the probability of the other. The test and the multiplication rule are $P(A\cap B)=P(A)\times P(B).$ If $A$ and $B$ are independent then so are $A'$ and $B$, $A$ and $B'$, and $A'$ and $B'$. For $P(A)=0.6$, $P(B)=0.5$: $P(A\cap B)=0.3$, $P(A\cup B)=0.8$ and $P(A'\cap B)=0.4\times0.5=0.2$. To show independence, check that $P(A\cap B)=P(A)P(B)$. Do not assume it unless the question states it or the context (for example separate tosses of a coin) makes it clear.

These ideas are often confused. If $A$ and $B$ are mutually exclusive and both have non-zero probability, they are not independent: $P(A\cap B)=0$ but $P(A)P(B)>0$. Knowing that $A$ has happened tells you $B$ cannot, so the events strongly affect each other. Example: $P(A)=0.4$, $P(B)=0.35$, mutually exclusive. Then $P(A)P(B)=0.14\ne0=P(A\cap B)$. Independent events are not mutually exclusive: if $P(A)=0.5$, $P(B)=0.2$ and $P(A\cup B)=0.6$, then $P(A\cap B)=0.1=P(A)P(B)$, so they are independent but not mutually exclusive.

Many questions use both rules. For events $C$ and $D$ that are mutually exclusive, and an event $E$ independent of each, $P(E\cap(C\cup D))=P(E\cap C)+P(E\cap D)=P(E)\big(P(C)+P(D)\big)$. Worked example. A pair of items is made, one on each of two machines, with defect probabilities $0.04$ and $0.06$ independently. Exactly one defective: $0.04\times0.94+0.96\times0.06=0.0952$ (the two cases are mutually exclusive, so add). No defective in three pairs: $(0.96\times0.94)^3=0.735$ (independent, so multiply). A useful pattern: 'and' means multiply when independent; 'or' means add when mutually exclusive.

The same ideas apply to random variables. For a discrete distribution, the outcomes $X=1$ and $X=2$ are mutually exclusive, so $P(X=1\text{ or }2)=P(X=1)+P(X=2)$. For independent repeated trials, such as three tosses of a coin or the trials of a binomial model, probabilities are multiplied. For a continuous distribution, events such as $X<2$ and $X>5$ cannot both happen, so they are mutually exclusive and $P(X<2\text{ or }X>5)=P(X<2)+P(X>5)$. Independence of trials is an assumption you should state: it justifies multiplying the probabilities.