Question

the venn diagram below shows the 11 students in mr. tran’s class. the diagram shows the memberships for the science club and the volleyball club. note that \linda\ and
achel\ are outside the circles since they are not members of either club. one student from the class is randomly selected. let a denote the event \the student is in the science club.\ let b denote the event \the student is in the volleyball club.\ (a) find the probabilities of the events below. write each answer as a single fraction. p(a) = p(b) = p(a and b) = p(a or b) = p(a) + p(b) − p(a and b) = (b) select the probability that is equal to p(a) + p(b) − p(a and b). ∘ p(a and b) ∘ p(a) ∘ p(a or b)

Answer:

Part (a)

Step 1: Find \( P(A) \)

• Explanation: Count students in Science Club (A). Science Club has Chang, Manuel, Kala, Felipe. Total: 4? Wait, no. Wait the Venn diagram: Science circle (only Science) has Chang, Manuel, Kala (3), and the overlap (Felipe). So total in A: 3 + 1 = 4? Wait no, let's count again. Wait the Venn diagram: Science circle (left) has Chang, Manuel, Kala (3), overlap (Felipe, 1), Volleyball circle (right) has Susan, Abdul, Austin, Carmen, Jessica (5) and overlap (Felipe, 1). Outside: Linda, Rachel (2). Total students: 3 (only Science) + 1 (overlap) + 5 (only Volleyball) + 2 (outside) = 3+1+5+2=11, which matches. So event A (Science Club) is students in Science circle: only Science (3) + overlap (1) = 4? Wait no: Chang, Manuel, Kala are only Science? Wait the diagram: Science circle has Chang, Manuel, Kala (left part) and Felipe (overlap). So total in A: 3 + 1 = 4? Wait no, wait the names: Science circle (including overlap) is Chang, Manuel, Kala, Felipe. So 4 students. Total students: 11. So \( P(A) = \frac{\text{Number in A}}{\text{Total}} = \frac{4}{11} \)? Wait no, wait: Wait the Science circle: Chang, Manuel, Kala (3) in only Science, and Felipe (1) in both. So total in A: 3 + 1 = 4. Yes. So \( P(A) = \frac{4}{11} \)? Wait no, wait: Wait the Volleyball circle: Susan, Abdul, Austin, Carmen, Jessica (5) in only Volleyball, and Felipe (1) in both. So total in B: 5 + 1 = 6. So \( P(B) = \frac{6}{11} \). \( P(A \text{ and } B) \) is the overlap: Felipe, so 1 student. So \( P(A \text{ and } B) = \frac{1}{11} \). \( P(A \text{ or } B) \) is students in A or B: only A (3) + only B (5) + both (1) = 3 + 5 + 1 = 9. So \( P(A \text{ or } B) = \frac{9}{11} \). Then \( P(A) + P(B) - P(A \text{ and } B) = \frac{4}{11} + \frac{6}{11} - \frac{1}{11} = \frac{4 + 6 - 1}{11} = \frac{9}{11} \). Wait let's recheck:

• Total students: 11 (Linda, Rachel, Chang, Manuel, Kala, Felipe, Susan, Abdul, Austin, Carmen, Jessica). So 11.

• Event A (Science Club): Chang, Manuel, Kala, Felipe. So 4 students. So \( P(A) = \frac{4}{11} \).

• Event B (Volleyball Club): Felipe, Susan, Abdul, Austin, Carmen, Jessica. So 6 students. So \( P(B) = \frac{6}{11} \).

• \( A \text{ and } B \): Felipe (only 1). So \( P(A \text{ and } B) = \frac{1}{11} \).

• \( A \text{ or } B \): students in A or B: Chang, Manuel, Kala, Felipe, Susan, Abdul, Austin, Carmen, Jessica. That's 9 students. So \( P(A \text{ or } B) = \frac{9}{11} \).

• \( P(A) + P(B) - P(A \text{ and } B) = \frac{4}{11} + \frac{6}{11} - \frac{1}{11} = \frac{4 + 6 - 1}{11} = \frac{9}{11} \).

Step 2: \( P(A) \)

• Explanation: Count in A: 4. Total: 11.

\( P(A) = \frac{4}{11} \)

Step 3: \( P(B) \)

• Explanation: Count in B: 6. Total: 11.

\( P(B) = \frac{6}{11} \)

Step 4: \( P(A \text{ and } B) \)

• Explanation: Count in both: 1. Total: 11.

\( P(A \text{ and } B) = \frac{1}{11} \)

Step 5: \( P(A \text{ or } B) \)

• Explanation: Count in A or B: 9. Total: 11.

\( P(A \text{ or } B) = \frac{9}{11} \)

Step 6: \( P(A) + P(B) - P(A \text{ and } B) \)

• Explanation: Substitute values: \( \frac{4}{11} + \frac{6}{11} - \frac{1}{11} = \frac{9}{11} \)

\( P(A) + P(B) - P(A \text{ and } B) = \frac{9}{11} \)

Part (b)

Step 1: Recall the principle of inclusion - exclusion

• Explanation: The formula for \( P(A \text{ or } B) \) is \( P(A) + P(B) - P(A \text{ and } B) \). So this is equal to \( P(A \text{ or } B) \).

Final Answers

(a)

\( P(A) = \boxed{\frac{4}{11}} \)
\( P(B) = \boxed{\frac{6}{11}} \)
\( P(A \text{ and } B) = \boxed{\frac{1}{11}} \)
\( P(A \text{ or } B) = \boxed{\frac{9}{11}} \)
\( P(A) + P(B) - P(A \text{ and } B) = \boxed{\frac{9}{11}} \)

(b)

The correct option is \( \boxed{P(A \text{ or } B)} \)