Question

For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1. (Hint: Apply de Morgan’s law and then the Bonferroni inequality). Derive below Results 1 to 4 from Axioms 1 to 3 given in Section 2.1.2 in the textbook.

Result 1: P (Ac) = 1 − P(A)

Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B)

Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)

Result 4: If B ⊂ A, thenA∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) and P (A) ? P(B).

Answer 1

1)

There are
A∪Ac = S and A∩Ac =∅,
Therefore, by Axiom 3,
P(A∪Ac) = P(A)+P(Ac) = 1,

since P(A∪Ac) = P(S) = 1, whence P(Ac) = 1−P(A).

2)

There is A∪(B∩Ac) = (A∪B)∩(A∪Ac) = A∪B, which is to say that A∪B can be expressed as the union of two disjoint sets. Therefore, according to axiom 3, there is

P(A∪B) = P(A)+P(B∩Ac).

But B = B∩(A∪Ac) = (B∩A)∪(B∩Ac) is also the union of two disjoint sets, so there is also

P(B) = P(B∩A)+P(B∩Ac) =⇒P(B∩Ac) = P(B)−P(B∩A).

Substituting the latter expression into the one above gives

P(A∪B) = P(A)+P(B)−P(A∩B).

3)

Consider
A = A∩(B∪Bc) = (A∩B)∪(A∩Bc).

The final expression denotes the union of disjoint sets, so there is

P(A) = P(A∩B)+P(A∩Bc).