Question

suppose that a certain mathematics class contains 26 students. of these, 14 are in band, 15 are in sports, and 7 are in neither band nor sports. suppose that a person is selected at random from the class. click true or false for each statement. the probability of selecting a person that is in sports and in band is \\( \frac { 5 } { 13 } \\). the probability of selecting a person that is not in sports and is in band is \\( \frac { 7 } { 11 } \\). a person in sports is selected. the probability that the person is also in band is \\( \frac { 2 } { 3 } \\). a person in band is selected. the probability that the person is also in sports is \\( \frac { 5 } { 14 } \\).

Explanation:

Step1: Calculate the number of students in both band and sports

Let \(n(B)\) be the number of students in band (\(n(B)=14\)), \(n(S)\) be the number of students in sports (\(n(S) = 15\)), and the total number of students \(n(T)=26\).
We know that \(n(B\cup S)=n(B)+n(S)-n(B\cap S)\), and the number of students in neither \(n(\text{neither})=7\). So \(n(B\cup S)=26 - 7=19\).
Then \(n(B\cap S)=n(B)+n(S)-n(B\cup S)=14 + 15-19=10\).

Step2: Calculate the probabilities

The probability of selecting a person in sports and in band \(P(B\cap S)=\frac{n(B\cap S)}{n(T)}=\frac{10}{26}=\frac{5}{13}\).
The probability of selecting a person not in sports and in band \(P(\overline{S}\cap B)=n(B)-n(B\cap S)=14 - 10 = 4\), so \(P(\overline{S}\cap B)=\frac{4}{26}=\frac{2}{13}
eq\frac{7}{11}\).
The probability of selecting a person in band and also in sports \(P(S|B)=\frac{n(B\cap S)}{n(B)}=\frac{10}{14}=\frac{5}{7}
eq\frac{5}{14}\).

Answer:

• The probability of selecting a person that is in sports and in band: True
• The probability of selecting a person that is not in sports and is in band: False
• A person in band is selected. The probability that the person is also in sports: False