Asked by j
Explain why P(A|D) and P(D|A) from the table below are not equal.
A 4-column table has 3 rows. The first column has entries A, B, total. The second column is labeled C with entries 6, 1, 7. The third column is labeled D with entries 2, 8, 10. The fourth column is labeled Total with entries 8, 9, 17.
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GPT-5 mini
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Compute both conditionals from the table (counts):
- P(A and D) = 2.
- P(D) = 10, so P(A | D) = 2/10 = 0.20 (20%).
- P(A) = 8, so P(D | A) = 2/8 = 0.25 (25%).
They are not equal because the condition changes the denominator — P(A|D) asks “of the 10 D cases, what fraction are A?” while P(D|A) asks “of the 8 A cases, what fraction are D?” In general P(X|Y) ≠ P(Y|X). By Bayes’ rule,
P(A|D) = P(D|A) · P(A) / P(D),
so P(A|D) = P(D|A) only when P(A) = P(D) (or in degenerate cases). Here P(A) = 8 and P(D) = 10, so the two conditionals differ.
- P(A and D) = 2.
- P(D) = 10, so P(A | D) = 2/10 = 0.20 (20%).
- P(A) = 8, so P(D | A) = 2/8 = 0.25 (25%).
They are not equal because the condition changes the denominator — P(A|D) asks “of the 10 D cases, what fraction are A?” while P(D|A) asks “of the 8 A cases, what fraction are D?” In general P(X|Y) ≠ P(Y|X). By Bayes’ rule,
P(A|D) = P(D|A) · P(A) / P(D),
so P(A|D) = P(D|A) only when P(A) = P(D) (or in degenerate cases). Here P(A) = 8 and P(D) = 10, so the two conditionals differ.
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