Questions: Part: 1 / 3 Part 2 of 3 (b) Find the probability that it was a pizza with one or more toppings, and it was not eaten at work. P (one or more topping pizza and not eaten at work) =

Part: 1 / 3

Part 2 of 3
(b) Find the probability that it was a pizza with one or more toppings, and it was not eaten at work.
P (one or more topping pizza and not eaten at work) =

Transcript text: Part: $1 / 3$ Part 2 of 3 (b) Find the probability that it was a pizza with one or more toppings, and it was not eaten at work. $P$ (one or more topping pizza and not eaten at work) $=$ $\square$ Save For Later Skip Part Check © 2025 McGraw Hill LLC. All Rights Reseved. Terms of Use

Solution

Solution Steps

Step 1: Define the Probabilities

Let $ P(A) $ be the probability that a pizza has one or more toppings, and let $ P(B) $ be the probability that a pizza was not eaten at work. From the given data, we have:

$ P(A) = 0.7 $
$ P(B) = 0.6 $

Step 2: Calculate the Joint Probability

To find the probability that a pizza has one or more toppings and was not eaten at work, we calculate the joint probability $ P(A \cap B) $. Assuming the events are independent, we use the formula: \[ P(A \cap B) = P(A) \cdot P(B) \]

Step 3: Substitute the Values

Substituting the values into the equation, we get: \[ P(A \cap B) = 0.7 \cdot 0.6 = 0.42 \]

Thus, the probability that it was a pizza with one or more toppings and it was not eaten at work is $ P(A \cap B) = 0.42 $.