To solve these questions, we'll use the conditional probability formula:

$P(A|C) = \frac{P(A \cap C)}{P(C)}$

$P(C|A) = \frac{P(A \cap C)}{P(A)}$

$P(B|E) = \frac{P(B \cap E)}{P(E)}$

Step-by-Step Solution:

a) $P(A|C)$

From the table:

• The frequency of $A \cap C = 8$
• The total frequency of $C = 8 + 6 = 14$

$P(A|C) = \frac{8}{14} = 0.57$

---

b) $P(C|A)$

From the table:

• The frequency of $A \cap C = 8$
• The total frequency of $A = 8 + 3 + 15 = 26$

$P(C|A) = \frac{8}{26} \approx 0.31$

---

c) $P(B|E)$

From the table:

• The frequency of $B \cap E = 5$
• The total frequency of $E = 15 + 5 = 20$

$P(B|E) = \frac{5}{20} = 0.20$

Final Results:

1. $P(A|C) = 0.57$
2. $P(C|A) = 0.31$
3. $P(B|E) = 0.20$

---

If you have any questions or need further clarification, let me know!

Here are 5 related questions you might consider:

1. How would the probabilities change if the values in the table were adjusted?
2. Can you derive a general formula for conditional probabilities?
3. What is the complement of $P(A|C)$?
4. How does the law of total probability apply to contingency tables?
5. What happens if you calculate $P(C|E)$?

Tip: Always double-check the total frequencies when calculating probabilities from a contingency table to avoid errors.