Question

1. in a recent survey of 100 people about their favorite fruit, 30 said they liked apples, 20 said they liked oranges, 15 said they liked grapes, and everyone else liked other fruits. a person from the survey is picked at random. a. what is the probability of picking someone who likes grapes or apples? 5. in a recent study of 80 people about music genres, 40 of them liked rap, 30 liked r&b, 20 like both rap and r&b. a person from the survey is picked at random. a. what is the probability that the person chosen likes rap or r&b? 6. at a school, 20% of students play soccer, 10% play basketball, and 5% play both. a. what is the probability a student plays soccer or basketball? b. what is the probability a student plays neither sport?

Explanation:

4.

Step1: Identify relevant values

Let \(A\) be the set of people who like apples (\(n(A)=30\)), \(G\) be the set of people who like grapes (\(n(G) = 20\)), and the total number of people \(n(T)=100\). Since there is no mention of overlap between liking apples and grapes, we assume they are mutually - exclusive.

Step2: Apply probability formula for mutually - exclusive events

The probability of an event \(E\) is \(P(E)=\frac{n(E)}{n(T)}\). For the event of liking apples or grapes (\(A\cup G\)), when \(A\) and \(G\) are mutually - exclusive, \(P(A\cup G)=P(A)+P(G)\).
\[P(A)=\frac{30}{100}\]
\[P(G)=\frac{20}{100}\]
\[P(A\cup G)=\frac{30 + 20}{100}=\frac{50}{100}=0.5\]

Step1: Identify relevant values

Let \(R\) be the set of people who like Rap (\(n(R)=40\)), \(B\) be the set of people who like R&B (\(n(B)=30\)), and the number of people who like both \(n(R\cap B)=20\), and the total number of people \(n(T) = 80\).

Step2: Apply the inclusion - exclusion principle for probability

The formula for \(P(R\cup B)\) is \(P(R\cup B)=P(R)+P(B)-P(R\cap B)\).
\[P(R)=\frac{40}{80}\]
\[P(B)=\frac{30}{80}\]
\[P(R\cap B)=\frac{20}{80}\]
\[P(R\cup B)=\frac{40 + 30- 20}{80}=\frac{50}{80}=\frac{5}{8}=0.625\]

Step1: Calculate the probability of playing soccer or basketball

Let \(S\) be the event of playing soccer (\(P(S)=0.2\)), \(B\) be the event of playing basketball (\(P(B)=0.1\)), and \(P(S\cap B)=0.05\).
Using the inclusion - exclusion principle \(P(S\cup B)=P(S)+P(B)-P(S\cap B)\).
\[P(S\cup B)=0.2 + 0.1-0.05=0.25\]

Step2: Calculate the probability of playing neither sport

The probability of the complement of an event \(E\) is \(P(\overline{E})=1 - P(E)\). Let \(E = S\cup B\), then \(P(\overline{S\cup B})=1 - P(S\cup B)\).
\[P(\overline{S\cup B})=1 - 0.25 = 0.75\]

Answer:

\(0.5\)

5.