Prove that (i) P(S|F) = P(F|F) = 1 and (ii) P(E′|F) = 1 − P(E|F), where P(F) ≠ 0.

1. 1 · Write what is given and what is to be proved. Given: events E and F in a sample space S, with P(F) ≠ 0. To prove: the two properties.
2. 2 · Draw the figure or sketch. Draw a rectangle S with overlapping circles E and F. Shade F, and mark the parts E ∩ F and E′ ∩ F.
3. 3 · Write the definition. P(A|F) = P(A ∩ F)/P(F) for any event A.
4. 4 · Prove (i).
   • P(S|F) = P(S ∩ F)/P(F) = P(F)/P(F) = 1
   • P(F|F) = P(F ∩ F)/P(F) = P(F)/P(F) = 1
5. 5 · Prove (ii).
   • S = E ∪ E′, and E ∩ E′ = φ
   • So E ∩ F and E′ ∩ F are disjoint, and their union is F
   • P(E|F) + P(E′|F) = P(F)/P(F) = 1
6. 6 · Check and box the answer. Hence P(E′|F) = 1 − P(E|F). Both results are proved.

Scoring tip: Quote the definition P(A|F) = P(A ∩ F)/P(F) at the start. Every later line uses it.