91Ó°ÊÓ

A set of four adults is selected. It is given that 40% of adults prefer to get their news online.

a.

Here, O denotes an adult who prefers online news, and D denotes an adult who prefers a source other than online news.

The probability of selecting an adult who prefers online news is computed below:

$$\begin{gathered} P\left(O\right)=40\% \\ =0.40 \end{gathered}$$

The probability of selecting an adult who prefers a source different from online is computed below:

$$\begin{gathered} P\left(D\right)=1-P\left(O\right) \\ =1-0.40 \\ =0.60 \end{gathered}$$

The probability of selecting the first three adults who prefer online news and the fourth adult who prefers a source different from online is computed below:

$$\begin{gathered} P\left(OOOD\right)=P\left(O\right)P\left(O\right)P\left(O\right)P\left(D\right) \\ =\left(0.4\right)\left(0.4\right)\left(0.4\right)\left(0.6\right) \\ =0.0384 \end{gathered}$$

Thus, $P\left(OOOD\right)=0.0384$.

b.

Consider the following arrangements of a selection of four adults when exactly three of them are denoted by O and one is denoted by D:

$$\begin{gathered} OOOD \\ OODO \\ ODOO \\ DOOO \end{gathered}$$

P(OOOD) is equal to 0.0384.

Now,

$$\begin{gathered} P\left(OODO\right)=P\left(O\right)P\left(O\right)P\left(D\right)P\left(O\right) \\ =\left(0.4\right)\left(0.4\right)\left(0.6\right)\left(0.4\right) \\ =0.0384 \end{gathered}$$

Thus, P(OODO) is equal to 0.0384.

$$\begin{gathered} P\left(ODOO\right)=P\left(O\right)P\left(D\right)P\left(O\right)P\left(O\right) \\ =\left(0.4\right)\left(0.6\right)\left(0.4\right)\left(0.4\right) \\ =0.0384 \end{gathered}$$

Thus, P(ODOO) is equal to 0.0384.

$$\begin{gathered} P\left(DOOO\right)=P\left(D\right)P\left(O\right)P\left(O\right)P\left(O\right) \\ =\left(0.6\right)\left(0.4\right)\left(0.4\right)\left(0.4\right) \\ =0.0384 \end{gathered}$$

Thus, P(DOOO) is equal to 0.0384.

c.

The probability ofgetting exactly three adults who prefer to get their news online and one adult who prefers a different news source is computed below:

$$\begin{gathered} \text{Prob}=P\left(OOOD\right)+P\left(OODO\right)+P\left(ODOO\right)+P\left(DOOO\right) \\ =0.0384+0.0384+0.0384+0.0384 \\ =0.1536 \end{gathered}$$

Therefore, the probability of getting exactly three adults who prefer to get their news online and one adult who prefers a different news source is equal to 0.1536.