 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/7/b/ii/solution

A \cap B

and

A \cap B^{c}

are disjoint and their union is

A

. Additivity of probability gives

P (A) = P (A \cap B) + P (A \cap B^{c}) .

Both conditioning events have positive probability because

0 < P (B) < 1

P (A) = P (A ∣ B) P (B) + P (A ∣ B^{c}) P (B^{c}) .

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