Let's break down the problem by analyzing the Venn diagram and solving each part:

Definitions:

• $A$ is the event "the student is in the Chess Club."
• $B$ is the event "the student is in the Science Club."

The Venn diagram shows the memberships:

• Chess only: Raina, Charlie, Elsa, Lucy, Pablo, Carlos (6 students)
• Both Chess and Science: Josh, Heather, Jina (3 students)
• Science only: Lisa, Keith, Kelko, Ann (4 students)
• Neither: Leila (1 student)

Total students: 14 (including Leila).

(a) Find the probabilities of the events:

1. $P(A)$: Probability the student is in the Chess Club.

   • Students in the Chess Club: $6 + 3 = 9$.
   • $P(A) = \frac{9}{14}$.

2. $P(B)$: Probability the student is in the Science Club.

   • Students in the Science Club: $4 + 3 = 7$.
   • $P(B) = \frac{7}{14}$.

3. $P(A \text{ and } B)$: Probability the student is in both clubs.

   • Students in both clubs: Josh, Heather, Jina (3 students).
   • $P(A \text{ and } B) = \frac{3}{14}$.

4. $P(A \mid B)$: Probability the student is in the Chess Club given they are in the Science Club.

   • This is calculated as $P(A \text{ and } B) / P(B) = \frac{3}{14} \div \frac{7}{14} = \frac{3}{7}$.

5. $P(B) \cdot P(A \mid B)$: This is just $P(A \text{ and } B)$.

   • $P(B) \cdot P(A \mid B) = \frac{7}{14} \times \frac{3}{7} = \frac{3}{14}$.

(b) Select the probability that is equal to $P(A \text{ and } B)$:

• The correct answer is $P(B) \cdot P(A \mid B)$ since it matches $P(A \text{ and } B)$.

Would you like further details or explanations?

Related questions:

1. How would the probabilities change if a new student joined only the Science Club?
2. How would you calculate $P(A \text{ or } B)$?
3. What is the probability that a student is in neither club?
4. What is $P(\text{not } A \mid B)$?
5. Can the Venn diagram be used to find conditional probabilities in other contexts?

Tip:

When working with Venn diagrams, always clearly define each section and count carefully to avoid over-counting or missing elements.