Chapter 13.10, Problem 39SR

a.

To determine

How many different breakfast of one bread as well as one beverage are possible?

Answer to Problem 39SR

  $9$ different breakfast can be served consisting of one bread as well as one beverage.

Explanation of Solution

As with the help of the following structure of the number of possible outcomes, as get the breakfast with any bread item and one beverage possible is as follows,

As in these there are $9$ different breakfasts can be served consisting of one bread and one beverage.

Toast − Coffee − Toast. Coffee

Milk − Toast. Milk

Orange Juice − Toast. Orange Juice

Muffin − Coffee − Muffin. Coffee

Milk − Muffin. Milk

Orange Juice - Muffin- Orange Juice

Bagel - Coffee − Bagel. Coffee

Milk − Bagel. Milk

Orange Juice − Bagel. Orange Juice

Conclusion:

Hence, $9$ different breakfast can be served consisting of one bread as well as one beverage.

b.

To determine

The probability that a customer chooses a bagel with orange juice if a bread and beverage are equally likely to be chosen.

Answer to Problem 39SR

The probability that a customer chooses a bagel with orange juice if a bread and beverage are equally likely to be chosen is $P\left(11.1\%\right)$ .

Explanation of Solution

Calculation:

In the fraction form the probability of choosing a bagel and orange juice if bread and a beverage is likely to be chosen is follows:

P (Bagel, Orange Juice) = $\frac{\text{No}\text{.}\text{of}\text{possible}\text{outcomes}}{\text{Total}\text{no}\text{.}\text{of}\text{outcomes}}=\frac{1}{9}$

In the percent form the probability of choosing a bagel and orange juice if bread and a beverage is likely to be chosen is as to convert from a fraction to the percent form so multiply the given fraction with the help of $100$ and finally place $\%$ at the end of the number.

  $\begin{array}{l}=\left(\frac{1}{9}\right)\times 100\%\\ =\left(\frac{100}{9}\right)\%\\ =11.1\%\end{array}$

Conclusion:

Hence The probability that a customer chooses a bagel with orange juice if a bread and beverage are equally likely to be chosen is $P\left(11.1\%\right)$ .