Q.State whether the following statement is True or False: If $A$ and $B'$ are independent events, then $P(A' \cup B) = 1 - P(A)\,P(B')$.

Concept: Event Independence — Two events are independent if the occurrence of one does not affect the probability of the other.

Reasoning:

1. Given $A$ and $B'$ are independent, so $P(A \cap B') = P(A)\,P(B')$.

2. We need $P(A' \cup B)$. By De Morgan’s law, $A' \cup B = (A \cap B')'$. …

Why This Approach Works

The problem asks whether a specific probability expression holds under the condition that $A$ and $B'$ are independent. Instead of blindly plugging into formulas, we need to connect the given expression to what independence tells us.

Independence of $A$ and $B'$ means:

Our target is $P(A' \cup B)$. The complement of this union is $A \cap B'$ — that’s the key insight. Because $A' \cup B$ and $A \cap B'$ are complements, their probabilities add to 1. So if we can find $P(A \cap B')$, we’re done.

Let’s walk through it step by step.

1. Recognize the complement relationship By De Morgan’s law:

This means $A' \cup B$ and $A \cap B'$ are complementary events. Therefore:

2. Use the given independence We are told $A$ and $B'$ are independent. So:

3. Substitute into the complement relation From step 1:

This is exactly the expression we needed to verify. …

Method: Evaluating a union with complements using De Morgan and independence

When you must find $P(\text{union involving complements})$ and are told some pair is independent, convert the union to a complement of an intersection so the independence product applies.

Steps

Step 1: Apply De Morgan to turn the union into a complemented intersection.

Step 2: Use the given independence on that intersection.

If $A$ and $B'$ are independent, then $P(A\cap B')=P(A)\,P(B')$. …

Common Mistakes

Mistake 1: Wrongly assuming $A'$ and $B$ are independent.

Why it's wrong: you are only told $A$ and $B'$ are independent; you must use that pair. Correct approach: De Morgan gives $A'\cup B=(A\cap B')'$, so $P(A'\cup B)=1-P(A\cap B')=1-P(A)P(B')$ using only the given independence.

Mistake 2: Mis-applying De Morgan. …