#### text

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)