Question

example: at a breakfast buffet, 93 people chose coffee and 47 people chose juice. 25 people chose both coffee and juice. if 130 people visited the buffet, what is the probability that...
a.) a person chose coffee?

b.) a person did not choose juice?

c.) a person chose both coffee and juice?

d.) a person chose either coffee or juice?

e.) a person chose neither coffee nor juice?

example: 150 college freshmen were interviewed. 85 were registered for a math class,
70 were registered for an english class, and 50 were registered for both math and
english. if one student is selected at random, what is the probability that that student...
a.) signed up only for a math class?

b.) signed up only for an english class?

c.) signed up for math or english?

d.) signed up neither for math nor english?

Explanation:

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First Breakfast Buffet Problem

Step1: Define total people

Total visitors $n(T) = 130$

Step2: Probability of choosing coffee

Number who chose coffee $n(C)=93$. Probability $P(C)=\frac{n(C)}{n(T)}$
$P(C)=\frac{93}{130}$

Step3: Probability of not choosing juice

Number who chose juice $n(J)=47$. Number who did not: $130-47=83$. Probability $P(\text{not }J)=\frac{83}{130}$

Step4: Probability of choosing both

Number who chose both $n(C\cap J)=25$. Probability $P(C\cap J)=\frac{25}{130}=\frac{5}{26}$

Step5: Probability of coffee or juice

Use addition rule: $P(C\cup J)=P(C)+P(J)-P(C\cap J)$
$P(C\cup J)=\frac{93}{130}+\frac{47}{130}-\frac{25}{130}=\frac{115}{130}=\frac{23}{26}$

Step6: Probability of choosing neither

$P(\text{neither})=1-P(C\cup J)$
$P(\text{neither})=1-\frac{23}{26}=\frac{3}{26}$

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Second College Freshmen Problem

Step1: Define total students

Total students $n(T)=150$

Step2: Probability of only Math

Only Math: $n(M)-n(M\cap E)=85-50=35$. Probability $P(\text{only }M)=\frac{35}{150}=\frac{7}{30}$

Step3: Probability of only English

Only English: $n(E)-n(M\cap E)=70-50=20$. Probability $P(\text{only }E)=\frac{20}{150}=\frac{2}{15}$

Step4: Probability of Math or English

Use addition rule: $P(M\cup E)=P(M)+P(E)-P(M\cap E)$
$P(M\cup E)=\frac{85}{150}+\frac{70}{150}-\frac{50}{150}=\frac{105}{150}=\frac{7}{10}$

Step5: Probability of neither

$P(\text{neither})=1-P(M\cup E)$
$P(\text{neither})=1-\frac{7}{10}=\frac{3}{10}$

Answer:

Breakfast Buffet:

a.) $\frac{93}{130}$
b.) $\frac{83}{130}$
c.) $\frac{5}{26}$
d.) $\frac{23}{26}$
e.) $\frac{3}{26}$

College Freshmen:

a.) $\frac{7}{30}$
b.) $\frac{2}{15}$
c.) $\frac{7}{10}$
d.) $\frac{3}{10}$