Q. 2.12 - A First Course in Probability

Question

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

Step-by-Step Solution

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 1 Given Information.

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

Step 2 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 $E$and$F$ 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

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