for any two events A and B ,P(AUB)=P(A)+P(B)-P(A intersection B) prove it
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for any two events A and B ,P(AUB)=P(A)+P(B)-P(A intersection B) prove itFor any sets A and B, we have the disjoint union
A∪B=(A−B)∪(A∩B)∪(B−A).
Then, by the axiom,
P(A∪B)=P(A−B)+P(A∩B)+P(B−A).
Since P(A−B)=P(A)−P(A∩B) and P(B−A)=P(B)−P(A∩B), the result follows.Here's how I'd do it:
Note that
(∗)P(A⋃B)=P(A⋃AcB)=P(A)+P(AcB)
Then since B=AB⋃AcB we get that
P(B)=P(AB)+P(AcB)
or equivalently
P(AcB)=P(B)−P(AB)
Plugging this into ∗ gives the desired result and thus completing the proof. Hope that helps!A rigorous proof of the statement P(A∪B)=P(A)+P(B)−P(A∩B) is performed using not only the axioms of probability theory but also the identities of set theory and the axioms of real numbers.
PROVE: For any arbitrary probability experiment consisting of a finite sample space, S, and events A and B s.t. A⊆S and B⊆S, the following is true: P(A∪B)=P(A)+P(B)−P(A∩B).
1. Let S be the finite sample space of some arbitrary probability experiment with events A and B s.t. A⊆S and B⊆S. First, we show P(A∪B)=P(A∪(B∩AC)).
A∪B=(A∪B)∩S by the identity law, where S, the sample space, is our universal set
=(A∪B)∩(A∪AC) by the negation law
=A∪(B∩AC) by the distributive law
Hence, A∪B=A∪(B∩AC); thus, we know
(1) P(A∪B)=P(A∪(B∩AC))
2. Second, we show the sets A and B∩AC are mutually exclusive or "disjoint" - or in other words, that A∩(B∩AC)=∅.
A∩(B∩AC)=A∩(AC∩B) by the commutative law
=(A∩AC)∩B by the associative law
=∅∩B by the negation law
=∅ by the domination law
Hence, A∩(B∩AC)=∅, implying that sets A and B∩AC are disjoint.
3. Since A and B∩AC are disjoint, then by the finite additvity axiom (axiom iii of modern probability theory) we may conclude P(A∪(B∩AC))=P(A)+P(B∩AC). Then, by substitution with equation (1), we have
(2) P(A∪B)=P(A)+P(B∩AC).
4. Next, we show P(B)=P((B∩A)∪(B∩AC)).
B=B∩S by the identity law
=B∩(A∪AC) by the negation law
=(B∩A)∪(B∩AC) by the distribution law
Hence, B=(B∩A)∪(B∩AC); thus, we know
(3) P(B)=P((B∩A)∪(B∩AC))
5. Now we show that sets B∩A and B∩AC are disjoint - or in other words, that (B∩A)∩(B∩AC)=∅.
(B∩A)∩(B∩AC)=B∩(A∩AC) by the distributive law
=B∩∅ by the negation law
=∅ by the domination law
Hence, (B∩A)∩(B∩AC)=∅, implying that sets B∩A and B∩AC are disjoint.