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: Understanding What Is to Be Proved:
"At least one of A or B occurs" is the event \(A\cup B\). We must show \(P(A\cup B)=1-P(A')P(B')\), given that A and B are independent, meaning \(P(A\cap B)=P(A)P(B)\).

Step 2: Use the Addition Theorem:
The probability of a union of two events is given by the addition theorem of probability.
\[ P(A\cup B)=P(A)+P(B)-P(A\cap B) \]

Step 3: Substitute the Independence Condition:
Since A and B are independent, replace \(P(A\cap B)\) with \(P(A)P(B)\).
\[ P(A\cup B)=P(A)+P(B)-P(A)P(B) \]

Step 4: Rewrite Using Complements:
Recall that \(P(A')=1-P(A)\) and \(P(B')=1-P(B)\), so their product can be expanded and compared with the expression above.
\[ P(A')P(B')=(1-P(A))(1-P(B))=1-P(A)-P(B)+P(A)P(B) \]
\[ 1-P(A')P(B')=P(A)+P(B)-P(A)P(B) \]
The right side here is identical to the expression obtained for \(P(A\cup B)\) in Step 3.

Final Answer:
Since both expressions equal \(P(A)+P(B)-P(A)P(B)\), they must be equal to each other, proving the result.
\[ \boxed{P(A\cup B)=1-P(A')P(B')} \]