Use the situation below to answer
1 answer:
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Answer:
a) see attachment
b) Pr(chicken sandwich, carrots, and a banana) = 1/12
c) The probability that Sol gets at least one of his favorite = 5/6
Step-by-step explanation:
a) A tree diagram is used to represent sample space.
We are told the lunch choices are sandwich, a vegetable and a fruit and each lunch combination are possible.
Find attached the tree diagram.
From the tree diagram, there are a total of 12 possible outcomes.
b) Sol's favorite lunch is a chicken sandwich, carrots, and a banana.
The probability that Sol gets his favorite lunch = Pr(chicken sandwich, carrots and a banana)
Pr(chicken sandwich, carrots, and a banana) = 1/12
Reason: From the tree diagram, the likelihood of having all three (chicken sandwich, carrots and a banana) occurs only once.
Pr(chicken sandwich, carrots and a banana) = [number of times(chicken sandwich, carrots and a banana) occurs]/(total number of possible outcomes)
Pr(chicken sandwich, carrots, and a banana) = 1/12
c) For Sol to get at least one of his favorite, we would have at least either chicken, carrot or banana in the possible outcome.
We have 10 of such possible outcome:
Chicken, carrot and apple
Chicken, carrot and banana
Chicken, spinach and apple
Chicken, spinach and banana
Hamburger, carrot and apple
Hamburger, carrot and banana
Hamburger, spinach and banana
Turkey, carrot and apple
Turkey, carrot and banana
Turkey, spinach and banana
The probability that Sol gets at least one of his favorite = (total number of at least one of Sol's favorite lunch)/(total possible outcome)
= 10/12 = 5/6
The probability that Sol gets at least one of his favorite = 5/6
