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).

By De morgan's law

which is Bonferroni’s inequality

Proof

If S is universal set then

Proof:

If S is a universal set then:

Which show A∪B can be expressed as union of two disjoint sets.

If A and (B∩Ac) are two disjoint sets then

B can be  expressed as:

If B is intersection of two disjoint sets then

Then (1) becomes

Proof:

If A and B are two disjoint sets then

Proof:

If B is subset of A then all elements of B lie in A so A ∩ B =B

where A and A ∩ Bc  are disjoint.

From axiom P(E)≥0

Therefore,

P(A)≥P(B)

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step-by-step explanation: