Question

describing conditional probabilities
explain why p(a|d) and p(d|a) from the table below are not equal.

	c	d	total
b	1	8	9
total	7	10	17

Explanation:

Step1: Recall Conditional Probability Formula

The formula for conditional probability is \( P(X|Y) = \frac{P(X \cap Y)}{P(Y)} \), or using counts from a table, \( P(X|Y) = \frac{\text{Number of cases where } X \text{ and } Y \text{ occur}}{\text{Number of cases where } Y \text{ occurs}} \).

Step2: Calculate \( P(A|D) \)

For \( P(A|D) \), the number of cases where \( A \) and \( D \) occur is 2 (from the table, cell A-D), and the number of cases where \( D \) occurs is 10 (total for column D). So \( P(A|D) = \frac{2}{10} = \frac{1}{5} \).

Step3: Calculate \( P(D|A) \)

For \( P(D|A) \), the number of cases where \( D \) and \( A \) occur is 2 (same cell A-D), and the number of cases where \( A \) occurs is 8 (total for row A). So \( P(D|A) = \frac{2}{8} = \frac{1}{4} \).

Step4: Compare the Two Probabilities

Since \( \frac{1}{5}
eq \frac{1}{4} \), the denominators (the total number of cases for the condition) are different (10 for \( D \), 8 for \( A \)), leading to different probabilities.

Answer:

\( P(A|D) = \frac{2}{10} = \frac{1}{5} \) and \( P(D|A) = \frac{2}{8} = \frac{1}{4} \). They differ because \( P(A|D) \) conditions on \( D \) (denominator 10) and \( P(D|A) \) conditions on \( A \) (denominator 8), so their denominators (total cases for the condition) are different, resulting in unequal probabilities.