Let's work through the problems based on the image you provided. I'll begin by interpreting and solving each part of the questions related to the Venn diagrams and the table provided.

1. First Venn Diagram (30 total outcomes)

• $A = 15$, $A \cap B = 5$, $B = 10$, total outcomes = 30.

a. Are $A$ and $B$ dependent or independent? Why?

• To determine whether $A$ and $B$ are independent, we check if $P(A \cap B) = P(A) \cdot P(B)$.

Let's calculate:

• $P(A) = \frac{15 + 5}{30} = \frac{20}{30} = \frac{2}{3}$
• $P(B) = \frac{10 + 5}{30} = \frac{15}{30} = \frac{1}{2}$
• $P(A \cap B) = \frac{5}{30} = \frac{1}{6}$

Now, check if $P(A \cap B) = P(A) \cdot P(B)$:

• $P(A) \cdot P(B) = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$
• Since $\frac{1}{6} \neq \frac{1}{3}$, $A$ and $B$ are dependent.

b. $P(A)$

$P(A) = \frac{20}{30} = \frac{2}{3}$.

c. $P(B|A)$

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{5}{30}}{\frac{20}{30}} = \frac{5}{20} = \frac{1}{4}$.

d. $P(A \cup B)$

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{2}{3} + \frac{1}{2} - \frac{1}{6} = \frac{4}{6} + \frac{3}{6} - \frac{1}{6} = \frac{6}{6} = 1$.

e. $P(A \cap B)$

$P(A \cap B) = \frac{5}{30} = \frac{1}{6}$.

2. Second Venn Diagram (32 total outcomes)

• $A = 8$, $A \cap B = 2$, $B = 16$, total outcomes = 32.

a. Are $A$ and $B$ dependent or independent? Why?

Let's calculate:

• $P(A) = \frac{8 + 2}{32} = \frac{10}{32} = \frac{5}{16}$
• $P(B) = \frac{16 + 2}{32} = \frac{18}{32} = \frac{9}{16}$
• $P(A \cap B) = \frac{2}{32} = \frac{1}{16}$

Now, check if $P(A \cap B) = P(A) \cdot P(B)$:

• $P(A) \cdot P(B) = \frac{5}{16} \cdot \frac{9}{16} = \frac{45}{256}$
• $P(A \cap B) = \frac{1}{16} = \frac{16}{256}$
• Since $\frac{45}{256} \neq \frac{16}{256}$, $A$ and $B$ are dependent.

b. $P(B)$

$P(B) = \frac{18}{32} = \frac{9}{16}$.

c. $P(A|B)$

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{2}{32}}{\frac{18}{32}} = \frac{2}{18} = \frac{1}{9}$.

d. $P(A \cup B)$

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{5}{16} + \frac{9}{16} - \frac{1}{16} = \frac{13}{16}$.

e. $P(A \cap B)$

$P(A \cap B) = \frac{2}{32} = \frac{1}{16}$.

3. Two-way Table (200 total outcomes)

a. Are $A$ and $B$ dependent or independent? Why?

To determine dependence, use the formula $P(A \cap B) = P(A) \cdot P(B)$.

• $P(A) = \frac{65}{200} = 0.325$
• $P(B) = \frac{100}{200} = 0.5$
• $P(A \cap B) = \frac{25}{200} = 0.125$

Now, check if $P(A \cap B) = P(A) \cdot P(B)$:

• $P(A) \cdot P(B) = 0.325 \cdot 0.5 = 0.1625$
• Since $0.125 \neq 0.1625$, $A$ and $B$ are dependent.

Would you like further clarification or details on any part? Here are some related questions to deepen your understanding:

1. How do we determine whether events are independent in probability theory?
2. Can you explain why the intersection $P(A \cap B)$ is important in determining dependence?
3. What is the significance of conditional probability $P(B|A)$?
4. How does the union formula $P(A \cup B)$ help in understanding combined probabilities?
5. What would change if $A$ and $B$ were independent events?

Tip: Always compare $P(A \cap B)$ with $P(A) \cdot P(B)$ to quickly check for independence.