Question

this question has two parts. use the information to answer part a and part b.
at lunch, there are 3 main dishes and 3 desserts. the main dish choices are a cheeseburger (c), hot dog (h), or salad (s). the dessert choices are a brownie (b), fruit cup (f), or ice cream (i). a customer chooses one main dish and one dessert for a meal.
part a
which list shows all possible combinations of 1 main dish and 1 dessert?
a. chs, bfi
b. cb, hf, si
c. cb, cf, ci, hb, hf, hi, sb, sf, si
d. ch, cs, cb, cf, ci, hs, hb, hf, hi, sb, sf, si, bf, bi, fi
part b
if a customer chooses 1 main dish and 1 dessert at random, what is the probability they will choose a cheeseburger and an ice cream?
a. \\(\frac{1}{15}\\)
b. \\(\frac{1}{9}\\)
c. \\(\frac{1}{3}\\)
d. \\(\frac{1}{2}\\)

Explanation:

Step1: Solve Part A (List all combinations)

Each main dish pairs with each dessert:

• Cheeseburger (C): $C+B=CB$, $C+F=CF$, $C+I=CI$
• Hot dog (H): $H+B=HB$, $H+F=HF$, $H+I=HI$
• Salad (S): $S+B=SB$, $S+F=SF$, $S+I=SI$

Combined list: $CB, CF, CI, HB, HF, HI, SB, SF, SI$

Step2: Solve Part B (Calculate total combinations)

Number of main dishes = 3, number of desserts = 3.
Total combinations: $3 \times 3 = 9$

Step3: Solve Part B (Find favorable outcome count)

Favorable outcome: Cheeseburger + Ice Cream (CI), count = 1

Step4: Solve Part B (Compute probability)

Probability = $\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{1}{9}$

Answer:

Part A: C. CB, CF, CI, HB, HF, HI, SB, SF, SI
Part B: B. $\frac{1}{9}$