Question

describing conditional probabilities

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

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$$\begin{tabular}{|c|c|c|c|} \\hline & c & d & total \\\\ \\hline a & 6 & 2 & 8 \\\\ \\hline b & 1 & 8 & 9 \\\\ \\hline total & 7 & 10 & 17 \\\\ \\hline \\end{tabular}$$

Explanation:

Calculate P(A|D)

Using the Conditional Probability Calculation knowledge point

$$ P(A|D) = \frac{P(A \cap D)}{P(D)} = \frac{2}{10} = 0.2 $$

Calculate P(D|A)

Using the Conditional Probability Calculation knowledge point

$$ P(D|A) = \frac{P(A \cap D)}{P(A)} = \frac{2}{8} = 0.25 $$

Compare the denominators

Using the Conditional Probability knowledge point

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Answer:

The conditional probabilities \(P(A|D)\) and \(P(D|A)\) are not equal because they restrict the sample space to different totals:

• \(P(A|D) = \frac{2}{10} = 0.2\) (the probability of \(A\) given that \(D\) has occurred, which uses the total of column \(D\), \(10\), as the denominator).
• \(P(D|A) = \frac{2}{8} = 0.25\) (the probability of \(D\) given that \(A\) has occurred, which uses the total of row \(A\), \(8\), as the denominator).

Since the condition restricts the sample space to different subgroups with different totals (\(10
eq 8\)), the resulting probabilities are different.