This exercise involves probability calculations and a weighted tree diagram related to a promotion scenario. Let’s address the questions step by step. Here's how we can tackle this:

1(a):

1. Probability $P(A)$: The probability that a customer buys the promoted camera.
   From the problem: $P(A) = 0.20$.

2. Probability $P(A \cap \overline{C})$: The probability that a customer buys the camera but does not buy the memory card.
   Given: 70% of customers who buy the camera also buy the memory card, so $P(C | A) = 0.70$. Thus:
   $P(A \cap \overline{C}) = P(A) \cdot P(\overline{C} | A) = 0.20 \cdot (1 - 0.70) = 0.20 \cdot 0.30 = 0.06$

1(b):
If a customer does not buy the camera ($\overline{A}$), we know:

• 60% of customers neither buy the camera nor the memory card, so:
  $P(\overline{A} \cap \overline{C}) = 0.60$
• $P(\overline{A}) = 1 - P(A) = 1 - 0.20 = 0.80$.
  Thus: $P(\overline{C} | \overline{A}) = \frac{P(\overline{A} \cap \overline{C})}{P(\overline{A})} = \frac{0.60}{0.80} = 0.75$

2. Construct the weighted tree diagram:
The tree includes the following branches:

• First decision: $P(A) = 0.20$, $P(\overline{A}) = 0.80$.
• Second decision (if $A$): $P(C | A) = 0.70$, $P(\overline{C} | A) = 0.30$.
• Second decision (if $\overline{A}$): $P(C | \overline{A}) = 0.25$, $P(\overline{C} | \overline{A}) = 0.75$.

3. Verifying $P(C) = 0.34$:
Using the law of total probability:
$P(C) = P(A) \cdot P(C | A) + P(\overline{A}) \cdot P(C | \overline{A})$ Substitute values:
$P(C) = (0.20 \cdot 0.70) + (0.80 \cdot 0.25) = 0.14 + 0.20 = 0.34$

4. Conditional probability $P(A | C)$:
Use Bayes’ theorem:
$P(A | C) = \frac{P(A \cap C)}{P(C)}$ We already know:

• $P(A \cap C) = P(A) \cdot P(C | A) = 0.20 \cdot 0.70 = 0.14$.
• $P(C) = 0.34$.
  Thus:
  $P(A | C) = \frac{0.14}{0.34} \approx 0.41$

5(a) Completing the table:

5(b) Expected benefit for 100 customers:
The expected benefit $E(X)$ is given by:
$E(X) = \sum (\text{Benefit} \cdot \text{Probability})$
Substitute values:
$E(X) = (0 \cdot 0.60) + (4 \cdot 0.28) + (30 \cdot 0.12) = 0 + 1.12 + 3.6 = 4.72 \, \text{€ per client}.$
For 100 customers:
$\text{Total benefit} = 100 \cdot 4.72 = 472 \, \text{€}.$

Do you need further clarifications or deeper delving into any of these steps?

Here are five related questions:

1. How would the calculations change if the probabilities were adjusted?
2. Can we verify $P(A \cup C)$ from the data?
3. How can the weighted tree diagram be drawn explicitly?
4. What happens to the benefit distribution if costs are included?
5. Can this method be generalized to other promotions?

Tip: For tree diagrams, always label each branch with probabilities and outcomes to simplify calculations later.