Question

unit 4 probability, random variables, and probability distributions
name: lucy luo
hw 3 homework - venn diagrams, unions, and intersections

1. at a liberal arts college in the midwest, 39% of freshman are enrolled in both a math course and an english

course, 11% are enrolled in a math course, but not an english course, and 63% are enrolled in an english
course.
a) construct a venn diagram to illustrate this
b) what is the probability that a freshman student
situation. be sure to label all sets.
selected at random is taking an english course but not
a mathematics course?
2
c) what is the probability a student is not enrolled in
either course?

1. at that same liberal arts college, 56% live in a campus residence hall, 62% participate in a campus meal

program and 42% do both.
a) construct a venn diagram to illustrate this
b) what is the probability that a randomly selected
situation. be sure to label all sets.
student either lives or eats on campus?
c) what is the probability that a randomly selected
person lives off campus and doesnt have a meal
program?

Explanation:

Step1: <Calculate the probability for part b of question 1>

The probability of taking an English course but not a mathematics course is given directly in the Venn - diagram as \(24\%\).

Step2: <Calculate the probability for part c of question 1>

The sum of all probabilities in a Venn - diagram (including the outside region) is \(100\%\). The sum of the probabilities of being in math only (\(11\%\)), in both (\(39\%\)), and in English only (\(24\%\)) is \(11\%+39\% + 24\%=74\%\). So the probability of not being in either course is \(100\%-(11\% + 39\%+24\%)=26\%\).

Step3: <Calculate the probability for part b of question 2>

Use the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Let \(A\) be the event of living in a campus residence hall (\(P(A) = 56\%\)) and \(B\) be the event of participating in a campus meal program (\(P(B)=62\%\)), and \(P(A\cap B) = 42\%\). Then \(P(A\cup B)=56\%+62\%-42\% = 76\%\).

Step4: <Calculate the probability for part c of question 2>

The sum of the probabilities of living on - campus only (\(P(A)-P(A\cap B)=56\% - 42\%=14\%\)), participating in the meal program only (\(P(B)-P(A\cap B)=62\% - 42\% = 20\%\)), and doing both (\(42\%\)) is \(14\%+20\%+42\%=76\%\). The probability of living off - campus and not having a meal program is \(100\%-(14\%+20\%+42\%)=24\%\).

Answer:

1. b) \(24\%\)

c) \(26\%\)

1. b) \(76\%\)

c) \(24\%\)