chapter 1

Chapter 2: Probabilityc. P(all H’s next to W’s | J & M together)= P(all H’s next to W’s – including J&M)/P(J&M together)6×1×4×1×2×16!5×2×1×4×3×2×16!48240= = = . 256. If P(B|A) > P(B), then P(B’|A) < P(B’).Proof by contradiction.Assume P(B’|A) ≥ P(B’).Then1 – P(B|A) ≥ 1 – P(B).- P(B|A) ≥ – P(B).P(B|A) ≤ P(B).This contradicts the initial condition, therefore P(B’|A) < P(B’).′P(A ∩ B)P(B)P(A′∩ B)P(B)P(A ∩ B)+ P(A′∩ B)P(B)P(B)P(B)57. P( A | B)+ P(A | B)= + == = 158.P[(A ∪ B)∩ C)P[(A ∩ C)∪ ( B ∩ C)]P(A ∪ B | C)==P(C)P(C)P(A ∩C)+ P(B ∩ C)− P(A ∩ B ∩ C)=P(C)= P(A|C) + P(B|C) – P(A ∩ B | C)73