Question 942689: Question 1: (4 points)
From historical data, the management of Dinny's Restaurant knows that the probability of a customer ordering a hamburger is 0.27, the probability of a customer ordering a coffee is 0.42, and probability of a customer ordering a hamburger as well as a coffee is 0.22.
Give all answers as decimals to 4 decimal places.
(a) What is the probability that customer will order a hamburger, a cup of coffee or both?
Solution:
(b) What is the probability that customer who orders a hamburger will also order a cup of coffee?
Solution:
(c) What is the probability that customer who orders a cup of coffee will also order a hamburger?
Solution:
(d) What is the probability that a customer orders neither a burger nor a coffee?
Solution:
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Given:
In a restaurant, probability of a customer
- ordering a hamburger, P(H) = 0.27
- ordering a coffee, P(C) = 0.42
- ordering a hamburger with a coffee, P(H∩C) = 0.22
Find:
(a) P(H∪C), i.e. a hamburger, a coffee or both
(b) P(C|H), i.e. orders a coffee given she orders a hamburger
(c) P(H|C), i.e. orders a hamburger given she orders a coffee
(d) P(~H∩~C), i.e. orders neither a hamburger nor a coffee.
Solution:
(a)Given P(H)=0.27, P(H∩C)=0.22, and P(C)=0.42, we calculate P(H∪C) using the relationship
P(H∪C) = P(H)+P(C)-P(H∩C) = 
(b)The required probability is a conditional probability, i.e.
P(C|H) = P(C∩H)/P(H) = 
(c) Again, it is a conditional probability, namely
P(H|C) = P(H∩C)/P(C) = 
(d) P(~H∩~C) = 1-P(H∪C) = 1-0.47 = 0.53
Answer:
(a) probability of ordering a hamburger, a coffee or both is 0.47
(b) probability of ordering a coffee given she orders a hamburger is 22/27
(c) probability of ordering a hamburger given she orders a coffee is 11/21
(d) probability of ordering neither a hamburger nor a coffee is 0.53
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