The problem involves a Venn diagram with two events: $A$ (students in the Tennis Club) and $B$ (students in the Computer Club). Let's analyze the data and compute the probabilities for the given questions.

Step 1: Count the members in each group based on the Venn diagram

• Total students: $10$ (including Karen, who is outside both circles)
• Tennis Club ($A$):
  • Members: Yolanda, Justin, Charmaine, Ann, Brian.
  • Total = 5 members.
• Computer Club ($B$):
  • Members: Alan, Elsa, Lashonda, Pablo, Ann, Brian.
  • Total = 6 members.
• Overlap ($A \cap B$):
  • Members: Ann, Brian.
  • Total = 2 members.
• Students outside both clubs (Karen):
  • Total = 1.

Step 2: Compute probabilities

We write probabilities as fractions of the total students, $n = 10$.

1. $P(A)$:

   • $P(A) = \frac{\text{Number in A}}{\text{Total students}} = \frac{5}{10} = \frac{1}{2}$.

2. $P(B)$:

   • $P(B) = \frac{\text{Number in B}}{\text{Total students}} = \frac{6}{10} = \frac{3}{5}$.

3. $P(A \cap B)$:

   • $P(A \cap B) = \frac{\text{Number in both A and B}}{\text{Total students}} = \frac{2}{10} = \frac{1}{5}$.

4. $P(A \cup B)$ (Union rule):

   • $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
   • $P(A \cup B) = \frac{1}{2} + \frac{3}{5} - \frac{1}{5} = \frac{5}{10} + \frac{6}{10} - \frac{2}{10} = \frac{9}{10}$.

5. Check formula $P(A) + P(B) - P(A \cap B)$:

   • Same as $P(A \cup B) = \frac{9}{10}$.

Step 3: Part (b) selection

• The probability equal to $P(A) + P(B) - P(A \cap B)$ is:
  • $P(A \cup B) = \frac{9}{10}$.

Would you like a detailed step-by-step explanation of each calculation? Or shall we proceed to analyze related concepts? Here are 5 related questions:

1. What is the probability that a randomly selected student is only in the Tennis Club?
2. What is the probability that a student is only in the Computer Club?
3. How would the probabilities change if Karen joined one of the clubs?
4. Can you express $P(A \cup B)$ in terms of conditional probabilities?
5. How does the Venn diagram help in visualizing disjoint and overlapping events?

Tip: Remember that probabilities of events must always sum up to $1$ across all possible outcomes.