Question

4-22. joan and jim are planning a dinner menu including a main dish and dessert. they have 4 main dish choices (steak, vegetable-cheese casserole, turkey burgers, and vegetarian lasagna) and 3 dessert choices (chocolate brownies, strawberry ice cream, and chocolate chip cookies.)
d. joan and jim would like to know how many different dinner menus they have to choose from. how many different menus are there?
e. what is the probability they pick a menu with meat?
f. what is the probability they pick a menu without chocolate?

Explanation:

Step1: Calculate total number of menus (d)

Use the fundamental counting principle. If there are \(m = 4\) main - dish choices and \(n=3\) dessert choices, the total number of menus is \(m\times n\).
\(4\times3 = 12\)

Step2: Calculate number of meat - based menus (e)

Assume steak and turkey burgers are meat - based. So there are \(2\) meat - based main dishes. With \(3\) dessert choices, the number of meat - based menus is \(2\times3=6\). The probability \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\), so \(P=\frac{6}{12}=\frac{1}{2}\)

Step3: Calculate number of non - chocolate menus (f)

Assume chocolate brownies and chocolate chip cookies are chocolate - based. So there is \(1\) non - chocolate dessert (strawberry ice cream). With \(4\) main - dish choices, the number of non - chocolate menus is \(4\times1 = 4\). The probability \(P=\frac{4}{12}=\frac{1}{3}\)

Answer:

d. \(12\)
e. \(\frac{1}{2}\)
f. \(\frac{1}{3}\)