EX 2.14 services of households

In a certain residential suburb,

• $60$ of all households get Internet service from the local cable company,
• $80$ get television service from that company,
• $50$ get both services from that company.

If a household is randomly selected,

• what is the probability that it gets at least one of these two services from the company
• what is the probability that it gets exactly one of these services from the company?

---

With

• $A$ = { gets Internet service }
• $B$ = { gets TV service }, the given information implies that $P\left(A\right)=.6$, $P\left(B\right)=.8$, and $P\left(A\cap B\right)=.5$. The foregoing proposition now yields

$$\begin{array}{rl}& P\left(\text{ subscribes to at least one of the two services }\right)\\ =& P\left(A\cup B\right)\\ =& P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\\ =& .6+.8-.5=.9\end{array}$$

---

The event that a household subscribes only to tv service can be written as $A^{\prime }\cap B$. Now Figure 2.4 implies that

$$\begin{array}{rl}.9& =P\left(A\cup B\right)=P\left(A\right)+P\left(A^{\prime }\cap B\right)\\ & =.6+P\left(A^{\prime }\cap B\right)\end{array}$$

from which $P\left(A^{\prime }\cap B\right)=.3$.

---

Similarly, $P\left(A\cap B^{\prime }\right)=P\left(A\cup B\right)-P\left(B\right)=.1$. This is all illustrated in Figure 2.5, from which we see that

$$\begin{array}{rl}P\left(\text{exactly one}\right)& =P\left(A\cap B^{\prime }\right)+P\left(A^{\prime }\cap B\right)\\ & =.1+.3=.4\end{array}$$

Figure 2.5 Probabilities for Example 2.14

---