11. Pizza Sales A pizza restaurant sold 24 cheese pizzasand 16 pizzas with one or more toppings. Twelve of thecheese pizzas were eaten at work, and 10 of the pizzaswith one or more toppings were eaten at work. If a pizzawas 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, andit was not eaten at work.c. It was a cheese pizza, or it was a pizza eaten atwork.

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Asked Oct 10, 2019
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Find the probability of each

11. 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.
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11. 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.

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Expert Answer

Step 1

11.a. The probability of it was a cheese pizza eaten at work is obtained below:

From the given information, number of cheese pizzas is 24, one or more topping pizzas is 16.  The 12 cheese pizzas were eaten at work and 10 one or more topping pizzas were eaten at work.

Total number of pizzas=24+16=40.

The required probability is,

Cheese pizza
P
eaten at work
12
40
0.30
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Cheese pizza P eaten at work 12 40 0.30

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Step 2

11.b. The probability of it was pizza with either or more toppings and it was not eaten at work is obtained below:

Let A be the event that pizza with either one or more toppings, B be the event that pizzas were not eaten at work.

The number of pizzas not eaten at work=40-(10+12)=18

The required probability is,

P(AB)P(A)+P(B) - P(AnB)
16
18
6
40
40
40
=0.40 0.45-0.15
-0.70
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P(AB)P(A)+P(B) - P(AnB) 16 18 6 40 40 40 =0.40 0.45-0.15 -0.70

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Step 3

11.c. The probability of it was cheese pizza, or it was a pizza eaten at work is obtained below:

Let A be the event that cheese pizza, B be the event that pizzas were eaten at work.

The number of pizzas eate...

P(AB)P(A)P(B) - P(AnB)
24 22 12
+
40 40 40
0.600.55-0.30
0.85
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P(AB)P(A)P(B) - P(AnB) 24 22 12 + 40 40 40 0.600.55-0.30 0.85

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