If A, B, C are three events associated with a random experiment, prove that
P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)
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Step-by-Step Solution
Step 1: Apply the Addition Theorem for Two Events
We can treat the union of three events, A∪B∪C, as the union of two events: event A and event (B∪C). This allows us to apply the addition theorem for two events, which states that P(X∪Y)=P(X)+P(Y)−P(X∩Y).
Step 2: Expand using the Addition Theorem
Applying the addition theorem for two events, P(X∪Y)=P(X)+P(Y)−P(X∩Y), where X=A and Y=(B∪C).
Step 3: Expand P(B∪C)
We apply the addition theorem again to expand the term P(B∪C). This is a standard result for the union of two events.
Step 4: Expand P(A∩(B∪C)) using Distributive Law
Using the distributive law of set theory, the intersection of event A with the union of events B and C is equivalent to the union of the intersection of A with B and the intersection of A with C.
Step 5: Apply Addition Theorem to P((A∩B)∪(A∩C))
We apply the addition theorem one more time to the expression P((A∩B)∪(A∩C)). Here, X=(A∩B) and Y=(A∩C).
Step 6: Simplify the last term
The intersection of (A∩B) and (A∩C) simplifies to A∩B∩C, as A is common to both intersections.
Step 7: Substitute and Combine Terms
Now, we substitute the expanded forms of P(B∪C) and P(A∩(B∪C)) back into the main equation from Step 2. Then, we distribute the negative sign and combine like terms to arrive at the final identity.