Question

1. pizza sales a pizza restaurant sold 24 cheese pizzas and 16 pizzas with one or more toppings. twelve of the cheese pizzas were eaten at work, and 10 of the pizzas with one or more toppings were eaten at work. if a pizza was selected at random, find the probability of each: a. it was a cheese pizza eaten at work. b. it was a pizza with either one or more toppings, and it was not eaten at work. c. it was a cheese pizza, or it was a pizza eaten at work.

Explanation:

Step1: Calculate total number of pizzas

Total number of pizzas \(n = 24+16=40\)

Step2: Solve part a

Number of cheese pizzas eaten at work \(n_a = 12\). Probability \(P(a)=\frac{n_a}{n}\)

$$P(a)=\frac{12}{40}=\frac{3}{10} = 0.3$$

Step3: Solve part b

Number of pizzas with one or more toppings \(n_{topping}=16\). Number of pizzas with one or more toppings eaten at work \(n_{topping - work}=10\). Number of pizzas with one or more toppings not eaten at work \(n_b=16 - 10=6\). Probability \(P(b)=\frac{n_b}{n}\)

$$P(b)=\frac{6}{40}=\frac{3}{20}=0.15$$

Step4: Solve part c

Let \(A\) be the event of cheese pizza and \(B\) be the event of pizza eaten at work. \(n(A) = 24\), \(n(B)=12 + 10=22\), \(n(A\cap B)=12\). By the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

$$P(A)=\frac{24}{40}, P(B)=\frac{22}{40}, P(A\cap B)=\frac{12}{40}$$

$$P(A\cup B)=\frac{24 + 22- 12}{40}=\frac{34}{40}=\frac{17}{20}=0.85$$

Answer:

a. \(0.3\)
b. \(0.15\)
c. \(0.85\)