The problem asks you to complete the table using the multiplication rule for probabilities, which is:

$P(A \cap B) = P(A) \cdot P(B|A)$ or equivalently, $P(A \cap B) = P(B) \cdot P(A|B)$

Let me fill in the missing values based on the given data and the multiplication rule.

First row:

• $P(A) = 0.8$, $P(B) = 0.5$, and $P(A|B) = 0.4$
  • $P(A \cap B) = P(B) \cdot P(A|B) = 0.5 \cdot 0.4 = 0.2$
  • $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2}{0.8} = 0.25$

Second row:

• $P(A) = 0.5$, $P(B) = 0.4$, and $P(B|A) = 0.2$
  • $P(A \cap B) = P(A) \cdot P(B|A) = 0.5 \cdot 0.2 = 0.1$
  • $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.4} = 0.25$

Third row:

• $P(A) = 0.8$, $P(A \cap B) = 0.5$, and $P(B|A) = 0.25$
  • $P(B) = \frac{P(A \cap B)}{P(A|B)} = \frac{0.5}{0.25} = 2$ → Check data (wrong value?)

Fourth row:

• $P(A) = 0.6$, $P(B) = 0.2$, and $P(A|B) = 0.5$
  • $P(A \cap B) = P(B) \cdot P(A|B) = 0.2 \cdot 0.5 = 0.1$
  • $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.6} \approx 0.17$

Fifth row:

• $P(A) = 0.4$, $P(B) = 0.5$, and $P(A \cap B) = 0.1$
  • $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.5} = 0.2$
  • $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.4} = 0.25$

Sixth row:

• $P(A) = 0.4$, $P(B) = 0.4$, and $P(A|B) = 0.6$
  • $P(A \cap B) = P(B) \cdot P(A|B) = 0.4 \cdot 0.6 = 0.24$
  • $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.24}{0.4} = 0.6$

Let me know if you want further clarifications or details.

Related Questions:

1. What is the significance of $P(A \cap B)$ in probability theory?
2. How is the conditional probability $P(B|A)$ calculated?
3. Can $P(A|B)$ ever be greater than $P(A)$?
4. What happens when $P(A \cap B) = 0$? What does this imply?
5. How would you interpret the multiplication rule in terms of independent events?

Tip:

Always check the consistency of the data in probability problems, as incorrect inputs can lead to contradictions in the results.