Exercise 2.17

Answers

(a): $P (B | A) = \frac{P (B) P (A | B)}{P (B) P (A | B) + P (B^{c}) P (A | B^{c})} = 1 ⟹ P (B^{c}) P (A | B^{c}) = 0$ .
Since $P (B^{c}) \neq 0$ by assumption, $P (A | B^{c}) = 0 ⟹ P (A^{c} | B^{c}) = 1$ .
(b): Let $A$ and $B$ be independent events. Then, $P (B | A) \approx 1 ⟹ P (B) \approx 1$ . Thus, $P (B^{c}) \approx 0$ , and so the term $P (A | B^{c})$ in the denominator in part $a$ may be large, implying $P (A^{c} | B^{c}) \approx 0$ .
For example, consider a deck of $52$ cards, where all but one of the cards are the Queen of Spades. Let $A$ be the event that the first turned card is a Queen of Spades, and let $B$ be the event that the second turned card is a Queen of Spades, where sampling is done with replacement. Then, $P (A) = P (B) \approx 1$ . Then, by independence, $P (A | B^{c}) \approx 1 ⟹ P (A^{c} | B^{c}) \approx 0$ .

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2021-12-05 00:00