91Ó°ÊÓ

Name the events:

$A$the event that a person buys a suit.

$B$the event that a person buys a shirt.

$C$the event that a person buys a tie.

Given:

$$\begin{gathered} P\left(A\right)=0.22 \\ P\left(B\right)=0.30 \\ P\left(C\right)=0.28 \\ P\left(AB\right)=0.11 \\ P\left(AC\right)=0.14 \\ P\left(BC\right)=0.10 \\ P\left(ABC\right)=0.06 \end{gathered}$$

In the terms of events $A,B$and $C$this is$P\left[(A∪B∪C{)}^c\right]$

$P\left[(A∪B∪C{)}^c\right]=1-P(A∪B∪C)\text{Proposition4.1}$

For future use, calculate $P(A∪B∪C)$using Proposition$4.4$(the first row of the following equation)

$$\begin{gathered} P(A∪B∪C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(AB)-P(AC)-P(BC)+P(ABC) \\ =0.22+0.30+0.28-0.11-0.14-0.10+0.06 \\ =0.51 \end{gathered}$$

Now

$$\begin{gathered} P\left[(A∪B∪C{)}^c\right]=1-0.51 \\ =0.49 \end{gathered}$$

$A∪B∪C$is the event that any item is bought.

$AC∪AC∪BC$is the event that any two events occurred.

So the wanted probability is$P\left[(A∪B∪C)(AC∪AC∪BC{)}^c\right]$.

Use the identity

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

For any events $E$and$F$.

$P(A∪B∪C)=P\left[\right(A∪B∪C\left)\right(AC∪AC∪BC\left)\right]+P\left[(A∪B∪C)(AC∪AC∪BC{)}^c\right]$

And since$(A∪B∪C)(AC∪AC∪BC)$is the event where any event happens, and any two events happen, it is equivalent to$(AC∪AC∪BC)$, that any two of these events happen.

$P(A∪B∪C)=P(AC∪AC∪BC)+P\left[(A∪B∪C)(AC∪AC∪BC{)}^c\right]$

And again using Proposition$4.4$,

$$\begin{gathered} P(AB∪AC∪BC)=P(AB)+P(AC)+P(BC) \\ -{\stackrel{Áåž}{P\left(ABAC\right)}}^{=P\left(ABC\right)}-{\stackrel{Áåž}{P\left(ACBC\right)}}^{=P\left(ABC\right)}-{\stackrel{Áåž}{P\left(ABBC\right)}}^{=P\left(ABC\right)}+P\left(ABC\right) \\ =0.11+0.14+0.10-2·0.06 \\ =0.23 \end{gathered}$$

Substitute this localid="1649316644779" $P(A∪B∪C)=0.51$into$\left(1\right)$

$0.51=0.23+P\left[(A∪B∪C)(AC∪AC∪BC{)}^c\right]$

This is equivalent to:

$P\left[(A∪B∪C)(AC∪AC∪BC{)}^c\right]=0.28$