Question

Prove that if \(A\) and \(B\) are independent events, then the probability of happening of at least one of \(A\) or \(B\) is \(\left[1-P(A')P(B')\right]\).

Hint:

Use the addition theorem P(A union B) = P(A) + P(B) - P(A)P(B), then compare with 1 - P(A')P(B').

Answer 1

Step 1: Start From the Complement Side:
Instead of starting from $P(A\cup B)$, start from the event "neither A nor B occurs," which is the complement of $A\cup B$.
By De Morgan's law, this complement event is $A'\cap B'$, so $P(A\cup B)=1-P(A'\cap B')$.

Step 2: Show A' and B' Are Also Independent:
Since A and B are independent, $P(A\cap B)=P(A)P(B)$, so use the addition theorem to compute $P(A'\cap B')$ through its complement.
\[ P(A'\cap B')=P((A\cup B)')=1-P(A\cup B)=1-[P(A)+P(B)-P(A)P(B)] \]
\[ =1-P(A)-P(B)+P(A)P(B)=(1-P(A))(1-P(B))=P(A')P(B') \]
This shows the complements $A'$ and $B'$ are themselves independent events.

Step 3: Substitute Back:
Now put this result for $P(A'\cap B')$ back into the expression from Step 1.
\[ P(A\cup B)=1-P(A'\cap B')=1-P(A')P(B') \]

Final Answer:
Working from the complement event and De Morgan's law also leads to the same required identity.
\[ \boxed{P(A\cup B)=1-P(A')P(B')} \]