Math proof required (Probability) [closed]

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Closed 8 years ago.

From Pr[E] = Pr[E|A].Pr[A] + Pr[E|A'].Pr[A']

how can we prove Pr[E] <= Pr[E|A] + Pr[A']

Answer 1

Pr[E] <= Pr[E|A] + Pr[A']

the left side can be replaced with the first line..

Pr[E|A].Pr[A] + Pr[E|A'].Pr[A'] <= Pr[E|A] + Pr[A']

hmm so lets subtract "Pr[E|A].Pr[A]" on both sides. On the right side, you can translate Pr[E|A] = Pr[E|A]*1 = Pr[E|A] (Pr[A] + Pr[A'])

Pr[E|A'].Pr[A'] <= Pr[E|A].Pr[A'] + Pr[A']

now we can put both sides into brackets in order to isolate Pr[A']

( Pr[E|A'] ) * Pr[A'] <= ( Pr[E|A] + 1 ) * Pr[A']

and divide by Pr[A']

Pr[E|A'] <= Pr[E|A] + 1

so.. if.. Pr[E|A] = 0 then both sides can be equal (if left side is 1) in all other cases the right side is larger, since it is more than 1, and left side can maximum be 1

Answer 2

Pr[E] <= Pr[E|A] is satisfied if Pr[E] is a subset of Pr[E|A]. This is proven with: Pr[E|A] = Pr[E] + Pr[A] - Pr[E&A].

So given any probability E and A, Pr[E] remains a subset of Pr[E|A].