problem 18

2.1.16 problem 18

$P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{P (B) - P (B \cap A^{c})}{P (B)} = 1 - \frac{P (B \cap A^{c})}{P (B)} .$

Note that $P (A^{c}) = P (A^{c} \cap B) + P (A^{c} \cap B^{c})$ . Hence, $P (B \cap A^{c}) = P (A^{c}) - P (A^{c} \cap B^{c}) = 0 - P (A^{c} \cap B^{c}) = - P (A^{c} \cap B^{c})$ .

Since probabilities are nonnegative, this implies $P (B \cap A^{c}) = P (A^{c} \cap B^{c}) = 0$ .

Thus, $P (A | B) = 1$ .

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