Question

1. A restaurant finds that 70% of its customers order an appetizer, 40% of its customers order a dessert, and 10% of customers order neither an appetizer nor a desert. (a) What is the probability that a customer orders both an appetizer and a dessert? (b) What is the probability that a customer orders dessert given that they ordered an appetizer?

2. You are the casting director for a short television commercial with three distinct characters. Twelve people have just auditioned for all three characters. How many possible ways are there for you to cast three of the twelve actors in these distinct roles?

Answer 1

Probability that the customer orders an appetizer=70/100=7/10=p(A)

Probability that the customer orders a dessert =40/100=4/10=p(B)

Probability that the customer orders neither appetizer not dessert is 10/100=1/10=1-p(AUB)

Probability that the customer orders both the appetizer and dessert is

P(A ∩ B)=p(A)+p(B)-p(AUB)

As p(AUB) is 9/10 (from given data)

P(A ∩ B)=7/10+4/10 -9/10

=2/10

B.

We need p(B/A)

P(B/A)=p(B ∩ A)/p(A)

=(2/10)/7/10

=2/7

2.

We need 3 distinct people for 3 distinct characters out of 12 people

This can be done in 12 c 3 ways

That is 120 ways.