Question

If A and B are two independent events, then the probability of occurrence of at least one of A and B is given by 1– P(A′) P(B′)

Answer 1

Is P(A′) P(B′) the probability that neither A nor B happens? I need to look at my Stat & Prob book. I wish we had a problem to test against this.

Answer 2

Yes

Answer 3

I was just re-reading Bayes' Theorem. No direct help accessible for me there.

Answer 4

that is correct because it includes all the possible outcomes except for the case where both A and B do not occur

Answer 5

I have to show it

Answer 6

So, the probability that A alone + prob of B alone + prob of A and B together + P(A′) P(B′) = 1.

How to prove that 1– P(A′) P(B′) --> I'm in a foreign land and don't know the language. :) But, that should be proved somewhere in a book or on a website.

Answer 7

Can you tell me what is P(at least one of A and B) ?
In My book it is given ( \(A \cup B\) ) Why is it so ?

Answer 8

I understood rest of the proof ! but li'l confused about it^

Answer 9

Atmost
For any two events occurrence of atmost one of the events implies the non occurrence of the event representing the intersection of those events.
At most one of the events "A" and "B" ⇒ The complimentary of intersection of the events i.e. (A ∩ B)c (A ∩ B)c = Ac ∪ Bc

For any three events occurrence of atmost two of the events implies the non occurrence of the event representing the intersection of those events.

At most two of the events "P", "Q" and "R" ⇒ The complimentary of intersection of the events i.e. (P ∩ Q ∩ R)c (P ∩ Q ∩ R)c = Pc ∪ Qc ∪ Rc

For any three events occurrence of atmost one of the events implies the non occurrence of the events representing the intersections of those events.

At most one of the events "P", "Q" and "R"
⇒ P(P ∩ Qc ∩ Rc) + P(Pc ∩ Q ∩ Rc) + P(Pc ∩ Qc ∩ R) + P(Pc ∩ Qc ∩ Rc)


http://www.futureaccountant.com/probability/study-notes/addition-theorem-probability-atleast-atmost-union-intersection.php

Answer 10

Now, what would be the union of the complements of sets A and B?

Answer 11

ok let P(A) = 1/m , P(B) = 1/n
P(A') = 1-1/m = (m-1)/m, P(B') = (n-1)/n

P(AB) = 1/mn
P(AB') = (n-1)/mn
P(A'B) = (m-1)/mn
P(A'B') = (m-1)(n-1)/mn = (mn-m-n+1)/mn

--> P(AB)+P(AB') +P(A'B) = 1-P(A'B')
1/mn + (n-1)/mn +(m-1)/mn = 1 - (mn-m-n+1)/mn
1/mn + (n-1)/mn +(m-1)/mn = mn-(mn-m-n+1)/mn
(m+n-1)/mn = (m+n-1)/mn
proof done

Answer 12

Ok Thanks Directrix & dumbcow

here's another one
P (at least one of A and B) = P(A ∪ B)
= P(A) + P(B) − P(A ∩ B)
= P(A) + P(B) − P(A) P(B)
(For independent events P(A ∩ B) =P(A) P(B) )

= P(A) + P(B) [1−P(A)]
= P(A) + P(B). P(A′)
= 1− P(A′) + P(B) P(A′)
= 1− P(A′) [1− P(B)]
= 1− P(A′) P (B′)