EX 2.26 reading habit

• The event whose probability is desired might be a union or intersection of other events,

  • the same could be true of the conditioning event.

• A news magazine publishes three columns entitled

  • “Art” $\left(A\right)$ ,
  • “Books” $\left(B\right)$ ,
  • “Cinema” $\left(C\right)$.

• Reading habits of a randomly selected reader with respect to these columns are

Read regularly	$A$	$B$	$C$	$A\cap B$	$A\cap C$	$A\cap B\cap C$
Probability	.14	.23	.37	.08	.09	.05

Figure 2.9 illustrates relevant probabilities.

Figure 2.9 Venn diagram for Example 2.26

Consider the following four conditional probabilities:

1. $P\left(A\mid B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{.08}{.23}=.348$
2. The probability that the selected individual regularly reads the Art column given that he or she regularly reads at least one of the other two columns is

$$\begin{array}{rl}P\left(A\mid B\cup C\right)& =\frac{P\left(A\cap \left(B\cup C\right)\right)}{P\left(B\cup C\right)}\\ & =\frac{.04+.05+.03}{.47}=\frac{.12}{.47}=.255\end{array}$$

3. $P\left(A\mid \text{reads at least one}\right)$ is equal to

$$\begin{array}{rl}P\left(A\mid A\cup B\cup C\right)& =\frac{P\left(A\cap \left(A\cup B\cup C\right)\right)}{P\left(A\cup B\cup C\right)}\\ & =\frac{P\left(A\right)}{P\left(A\cup B\cup C\right)}=\frac{.14}{.49}=.286\end{array}$$

4. The probability that the selected individual reads at least one of the first two columns given that he or she reads the Cinema column is

$$P\left(A\cup B\mid C\right)=\frac{P\left(\left(A\cup B\right)\cap C\right)}{P\left(C\right)}=\frac{.04+.05+.08}{.37}=.459$$