# Properties of Conditional Probability

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Let E and F be events of a sample space S of an experiment, then we have
Property :  P(S|F) = P(F|F) = 1
We know that
P(S|F) = (P(S ∩ F))/(P(F)) = (P(F))/(P(F)) = 1

Also  P(F|F) = (P(F ∩ F))/(P(F)) = (P(F))/(P(F)) = 1
Thus  P(S|F) = P(F|F) = 1

Property :  If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0, then
P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)
In particular, if A and B are disjoint events, then
P((A∪B)|F) = P(A|F) + P(B|F)
We have
P((A∪B)|F) = (P[(A ∪ B) ∩ F]) /(P(F))

= (P[(A ∩ F )∪ (B ∩ F)])/ (P(F))
(by distributive law of union of sets over  intersection)

= (P(A ∩ F) + P (B ∩ F) - P(A ∩ B ∩ F))/(P(F))

= (P(A ∩ F))/(P(F)) + (P (B ∩ F)) / (P(F)) - (P[(A ∩ B ∩ F)]) /(P(F))

= P(A|F) + P(B|F) – P((A∩B)|F)
When A and B are disjoint events, then
P((A ∩ B)|F) = 0
⇒ P((A ∪ B)|F) = P(A|F) + P(B|F)

Property : P(E′|F) = 1 − P(E|F)
From first Property , we know that P(S|F) = 1
⇒ P(E ∪ E′|F) = 1            since  S = E ∪ E′
⇒ P(E|F) + P (E′|F) = 1    since E and E′ are disjoint events
Thus, P(E′|F) = 1 − P(E|F)

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