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

Two events are mutually exclusive if they cannot both happen in the same trial, so $P(A\cap B)=0$. On a Venn diagram the circles do not overlap. The addition law simplifies to $P(A\cup B)=P(A)+P(B).$ For example, if $P(A)=0.3$ and $P(B)=0.4$ are mutually exclusive then $P(A\cup B)=0.7$ and $P(\text{neither})=1-0.7=0.3$. The outcomes of one trial, for example the faces of a die, are mutually exclusive, and their probabilities add to 1.

Events are independent if the occurrence of one does not change the probability of the other. Formally, $P(A\cap B)=P(A)P(B),\qquad P(B\mid A)=P(B),\qquad P(A\mid B)=P(A).$ Any one of these statements being true means all of them are. To test for independence, calculate $P(A)P(B)$ and compare it with $P(A\cap B)$; they must be equal. For independent events the union is $P(A\cup B)=P(A)+P(B)-P(A)P(B)$. If $A$ and $B$ are independent then so are $A$ and $B'$, $A'$ and $B$, and $A'$ and $B'$.

These ideas are different, and usually opposite. If $A$ and $B$ are mutually exclusive and both have non-zero probability, then $P(A\cap B)=0$ but $P(A)P(B)>0$, so they are not independent: if $A$ happens you know $B$ cannot, which changes the probability of $B$ to $0$. Worked example: $P(A)=0.5$ and $P(A\cup B)=0.8$. Mutually exclusive needs $0.8=0.5+P(B)$ so $P(B)=0.3$. Independent needs $0.8=0.5+P(B)-0.5P(B)$ so $P(B)=0.6$. The two conditions give different values, so they cannot both hold.

On a Venn diagram place $P(A\cap B)$ in the overlap first, then fill $P(A)-P(A\cap B)$ and $P(B)-P(A\cap B)$, and the outside region is $1-P(A\cup B)$. For mutually exclusive events the overlap is $0$. On a tree diagram multiply along branches and add between branches. For independent events the second set of branches carries the same probabilities whichever first branch was taken, so $P(C\cap D')=0.6\times0.75$. If the second-stage probabilities differ between branches, the events are not independent.

The possible values of a discrete random variable are mutually exclusive outcomes, so $P(X=x)$ values add to $1$ and $P(X\le 3)$ is found by adding the separate probabilities. Separate observations of a variable are normally taken to be independent, so $P(X_1=a\text{ and }X_2=b)=P(X_1=a)P(X_2=b)$. For a continuous random variable, probability is represented by area under the curve, so $P(a<X<b)$ is the area between $a$ and $b$, and the total area is $1$. Single values have no area, so $P(X=a)=0$. No formal use of probability density functions is needed at this stage.