Q.State whether the following statement is True or False: If $A$ and $B$ are independent, then $P(\text{exactly one of } A, B \text{ occurs}) = P(A)\,P(B') + P(B)\,P(A')$.

Concept: Event Independence — For independent events $A$ and $B$, $P(A \cap B) = P(A)P(B)$.

Reasoning:

1. "Exactly one of $A, B$ occurs" means $(A \cap B') \cup (A' \cap B)$, a disjoint union.
2. So $P(\text{exactly one}) = P(A \cap B') + P(A' \cap B)$. …

Why This Works: The Logic of "Exactly One"

When two events are independent, knowing that one happens tells you nothing about whether the other happens. This makes the "exactly one" case particularly clean: it's simply the union of two mutually exclusive possibilities — $A$ happens and $B$ does not, or $B$ happens and $A$ does not.

The key insight: independence lets us multiply probabilities for intersections. So $P(A \cap B') = P(A) P(B')$ and $P(B \cap A') = P(B) P(A')$. Since these two events can't happen at the same time (they're disjoint), we just add them.

Step-by-Step

1. Define the event "exactly one occurs" Exactly one of $A$ or $B$ occurs means: $A$ occurs and $B$ does not, or $B$ occurs and $A$ does not. In set notation:

2. Check if these two pieces overlap Can $A \cap B'$ and $B \cap A'$ happen together? That would require $A$ and $B$ both to occur and not occur simultaneously — impossible. So they are mutually exclusive (disjoint). Therefore:

3. Use independence to break down each intersection …

Method: Probability that exactly one of two independent events occurs

"Exactly one" of $A,B$ means one happens and the other does not. Split it into disjoint pieces and use independence on each.

Steps

Step 1: Write the event as a disjoint union.

and these two pieces cannot happen together, so their probabilities add.

Step 2: Apply independence to each intersection.

If $A,B$ are independent, so are $A,B'$ and $A',B$. Hence

$P(A\cap B')=P(A)P(B'),\qquad P(A'\cap B)=P(A')P(B).$ …

Common Mistakes

Mistake 1: Including the "both occur" case in "exactly one."

Why it's wrong: "exactly one" excludes $A\cap B$; it is $(A\cap B')\cup(A'\cap B)$ only. Correct approach: keep the two disjoint pieces and drop $A\cap B$.

Mistake 2: Not using that independence transfers to complements. …