The Correct Answer and Explanation is:
We are asked to prove the formula for the union of three events using the formula for the union of two events:P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) – P(A \cap B)P(A∪B)=P(A)+P(B)−P(A∩B)
We want to show that:P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(A∩C)+P(A∩B∩C)P(A \cup B \cup C) = P(A) + P(B) + P(C) – P(A \cap B) – P(B \cap C) – P(A \cap C) + P(A \cap B \cap C)P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(A∩C)+P(A∩B∩C)
Step-by-step Proof:
Let us define:P(A∪B∪C)=P((A∪B)∪C)P(A \cup B \cup C) = P((A \cup B) \cup C)P(A∪B∪C)=P((A∪B)∪C)
Apply the two-event union formula:P((A∪B)∪C)=P(A∪B)+P(C)−P((A∪B)∩C)P((A \cup B) \cup C) = P(A \cup B) + P(C) – P((A \cup B) \cap C)P((A∪B)∪C)=P(A∪B)+P(C)−P((A∪B)∩C)
Now substitute P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) – P(A \cap B)P(A∪B)=P(A)+P(B)−P(A∩B):P(A∪B∪C)=[P(A)+P(B)−P(A∩B)]+P(C)−P((A∪B)∩C)P(A \cup B \cup C) = [P(A) + P(B) – P(A \cap B)] + P(C) – P((A \cup B) \cap C)P(A∪B∪C)=[P(A)+P(B)−P(A∩B)]+P(C)−P((A∪B)∩C)
Now we simplify:P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P((A∪B)∩C)P(A \cup B \cup C) = P(A) + P(B) + P(C) – P(A \cap B) – P((A \cup B) \cap C)P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P((A∪B)∩C)
Next, we expand (A∪B)∩C(A \cup B) \cap C(A∪B)∩C using the distributive property:(A∪B)∩C=(A∩C)∪(B∩C)(A \cup B) \cap C = (A \cap C) \cup (B \cap C)(A∪B)∩C=(A∩C)∪(B∩C)
So:P((A∪B)∩C)=P((A∩C)∪(B∩C))=P(A∩C)+P(B∩C)−P(A∩B∩C)P((A \cup B) \cap C) = P((A \cap C) \cup (B \cap C)) = P(A \cap C) + P(B \cap C) – P(A \cap B \cap C)P((A∪B)∩C)=P((A∩C)∪(B∩C))=P(A∩C)+P(B∩C)−P(A∩B∩C)
Now substitute this back:P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−[P(A∩C)+P(B∩C)−P(A∩B∩C)]P(A \cup B \cup C) = P(A) + P(B) + P(C) – P(A \cap B) – [P(A \cap C) + P(B \cap C) – P(A \cap B \cap C)]P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−[P(A∩C)+P(B∩C)−P(A∩B∩C)]
Simplify the last expression:P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)P(A \cup B \cup C) = P(A) + P(B) + P(C) – P(A \cap B) – P(A \cap C) – P(B \cap C) + P(A \cap B \cap C)P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)
Final Identity Proven:
P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(A∩C)+P(A∩B∩C)P(A \cup B \cup C) = P(A) + P(B) + P(C) – P(A \cap B) – P(B \cap C) – P(A \cap C) + P(A \cap B \cap C)P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(A∩C)+P(A∩B∩C)
Explanation):
This problem involves the inclusion and exclusion principle in probability, which helps us compute the probability of the union of multiple events while accounting for overlaps. The principle is based on the idea that if we simply added the probabilities of all individual events, we would overcount the parts where events overlap. Thus, we must subtract those overlaps, but if we subtract too much, we also need to add back the shared overlap among all three events.
To start, we rewrite P(A∪B∪C)P(A \cup B \cup C)P(A∪B∪C) as P((A∪B)∪C)P((A \cup B) \cup C)P((A∪B)∪C). This allows us to apply the known formula for the union of two sets. We substitute the formula for P(A∪B)P(A \cup B)P(A∪B) into this expression. Then, we handle (A∪B)∩C(A \cup B) \cap C(A∪B)∩C using the distributive property, which becomes (A∩C)∪(B∩C)(A \cap C) \cup (B \cap C)(A∩C)∪(B∩C).
We then apply the two-set union formula again to simplify P((A∩C)∪(B∩C))P((A \cap C) \cup (B \cap C))P((A∩C)∪(B∩C)), which gives us the sum of the individual intersections minus the three-way intersection. Once we substitute everything back and combine like terms, we arrive at the final identity.
This formula is extremely useful in real-world problems, such as computing the chance that at least one of several events happens. Without subtracting the intersections correctly, we would double count overlapping probabilities and end up with an incorrect, overly large value.
