Question

Show that the probability that exactly one of the events$EorF$occurs equals $P\left(E\right)+P\left(F\right)-2P\left(EF\right).$

Short answer

Note that the event in which only one$E$,$F$occurs is$P\left((E\cup F)(EF{)}^c\right)$

Use Axiom$3$for $\left.(E\cup F)(EF{)}^c\right)$&$E\cup F$and Proposition$4.3$.

Step-by-step solution

Given Information.

The probability that exactly one of the events$EorF$occurs equals$P\left(E\right)+P\left(F\right)-2P\left(EF\right).$.

Explanation.

For any events $E$and$F$:

$P\left((E\cup F)(EF{)}^c\right)=P\left(E\right)+P\left(F\right)-2P\left(EF\right)$

Where the left-hand side is the event where precisely one event $E$&$F$occurs.

Because$(E\cup F)(EF{)}^c$is an event that occurs if$E$or $F$occur and not both $E$andlocalid="1649246913116" $F$occur.

Firstly note that

$$\begin{gathered} (E\cup F)(EF{)}^c\cap EF=\varnothing \\ (E\cup F\left)\right(EF{)}^c\cup EF=E\cup F \end{gathered}$$

Therefore

$$\begin{gathered} P(E\cup F)=P\left((E\cup F)(EF{)}^c\right)+P\left(EF\right) \\ P\left((E\cup F)(EF{)}^c\right)=P(E\cup F)-P\left(EF\right) \\ =P\left(E\right)+P\left(F\right)-P(EF)-P(EF) \\ =P\left(E\right)+P\left(F\right)-2P(EF) \end{gathered}$$