Respuesta :
Answer:
For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.
By De morgan's law
[tex](A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq 1-P(A)+1-P(B)\\\\-P(A\cap B)\leq 1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1[/tex]
which is Bonferroni’s inequality
Result 1: P (Ac) = 1 − P(A)
Proof
If S is universal set then
[tex]A\cup A^{c}=S\\\\P(A\cup A^{c})=P(S)\\\\P(A)+P(A^{c})=1\\\\P(A^{c})=1-P(A)[/tex]
Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)
Proof:
If S is a universal set then:
[tex]A\cup(B\cap A^{c})=(A\cup B) \cap (A\cup A^{c})\\=(A\cup B) \cap S\\A\cup(B\cap A^{c})=(A\cup B)[/tex]
Which show A∪B can be expressed as union of two disjoint sets.
If A and (B∩Ac) are two disjoint sets then
[tex]P(A\cup B) =P(A) + P(B\cap A^{c})---(1)\\[/tex]
B can be expressed as:
[tex]B=B\cap(A\cup A^{c})\\[/tex]
If B is intersection of two disjoint sets then
[tex]P(B)=P(B\cap(A)+P(B\cup A^{c})\\P(B\cup A^{c}=P(B)-P(B\cap A)[/tex]
Then (1) becomes
[tex]P(A\cup B) =P(A) +P(B)-P(A\cap B)\\[/tex]
Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)
Proof:
If A and B are two disjoint sets then
[tex]A=A\cap(B\cup B^{c})\\A=(A\cap B) \cup (A\cap B^{c})\\P(A)=P(A\cap B) + P(A\cap B^{c})\\[/tex]
Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc)
Proof:
If B is subset of A then all elements of B lie in A so A ∩ B =B
[tex]A =(A \cap B)\cup (A\cap B^{c}) = B \cup ( A\cap B^{c})[/tex]
where A and A ∩ Bc are disjoint.
[tex]P(A)=P(B\cup ( A\cap B^{c}))\\\\P(A)=P(B)+P( A\cap B^{c})[/tex]
From axiom P(E)≥0
[tex]P( A\cap B^{c})\geq 0\\\\P(A)-P(B)=P( A\cap B^{c})\\P(A)=P(B)+P(A\cap B^{c})\geq P(B)[/tex]
Therefore,
P(A)≥P(B)
