Q. 2.2 - A First Course in Probability

Question

A customer visiting the suit department of a certain store will purchase a suit with a probability of$.22$, a shirt with a probability of$.30$, and a tie with a probability$.28$. The customer will purchase both a suit and a shirt with probability $.11$, both a suit and a tie with probability$.14$, and both a shirt and a tie with probability$.10$. A customer will purchase all$3$ items with a probability of$.06$. What is the probability that a customer purchases

$\left(a\right)$none of these items?

$\left(b\right)$exactly$1$ of these items?

Step-by-Step Solution

Answer

Use Propositions $4.1$and$4.4$.

$$\begin{gathered} \text{a) }P\left[(A\cup B\cup C{)}^c\right]=0.49 \\ \text{ b) }P\left[(A\cup B\cup C)(AC\cup AC\cup BC{)}^c\right]=0.28 \end{gathered}$$

Step 1 Given Information.

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:

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

 

Step 2 Part (a) Explanation.

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

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

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

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

Now

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

Step 3 Part (b) Explanation.

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

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

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

Use the identity

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

For any events $E$and$F$.

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

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

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

And again using Proposition$4.4$,

$$\begin{gathered} P(AB\cup AC\cup BC)=P(A B)+P(A C)+P(B C) \\ -{\overbrace{P\left(ABAC\right)}}^{=P\left(ABC\right)}-{\overbrace{P\left(ACBC\right)}}^{=P\left(ABC\right)}-{\overbrace{P\left(ABBC\right)}}^{=P\left(ABC\right)}+P\left(ABC\right) \\ =0.11+0.14+0.10-2·0.06 \\ =0.23 \end{gathered}$$

Substitute this $P(A\cup B\cup C)=0.51$into$\left(1\right)$

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

This is equivalent to:

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