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

State whether the following statement is True or False: If A and B are independent, then P(exactly one of A, B

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The statement is True. For independent events, the probability that exactly one occurs is the sum of the probabilities of $A$ alone and $B$ alone, which matches the given expression.

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.

Watch out

A common mistake is to think "exactly one" means $P(A \cup B) - P(A \cap B)$ . That's also correct, but it's a different path. The given expression is simpler and directly uses independence.

Step-by-Step

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:

(A \cap B') \cup (B \cap A')

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:

P(\text{exactly one}) = P(A \cap B') + P(B \cap A')

Use independence to break down each intersection …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
$P(A) = \dfrac{2}{5},\ P(B) = \dfrac{1}{3}$
›Reveal solutionSolution

For independent events $P(A)=\tfrac25,\ P(B)=\tfrac13$ , the four quantities evaluate to $\tfrac{11}{15},\ \tfrac25,\ \tfrac35,\ \tfrac23$ , giving the match A–II, B–IV, C–III, D–I (option D).

Since $A$ and $B$ are independent, $P(A\cap B)=P(A)\,P(B)=\dfrac25\cdot\dfrac13=\dfrac{2}{15}$ , and each complement/conditional simplifies via independence.

List-I A) $P(\overline{A}\cup B)$
$P(\overline A\cup B)=P(\overline A)+P(B)-P(\overline A)P(B)=\dfrac35+\dfrac13-\dfrac35\cdot\dfrac13=\dfrac{9+5-3}{15}=\dfrac{11}{15}\quad(\text{II}).$
List-I B) $P(A\mid B)$ (independence $\Rightarrow$ conditioning has no effect)
$P(A\mid B)=P(A)=\dfrac25\quad(\text{IV}).$
List-I C) $P(A\cup B)$
$P(A\cup B)=P(A)+P(B)-P(A)P(B)=\dfrac25+\dfrac13-\dfrac{2}{15}=\dfrac{6+5-2}{15}=\dfrac{9}{15}=\dfrac35\quad(\text{III}).$
List-I D) $P(\overline{B}\mid A)$

$P(\overline B\mid A)=P(\overline B)=1-\dfrac13=\dfrac23\quad(\text{I}).$ …
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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 understanding — Event Independence

Event Independence

From Conditional Probability to a Clean Test

A Quick Check

Three or More Events

Why This Works: The Logic of "Exactly One"

Step-by-Step

Method: Probability that exactly one of two independent events occurs

Steps

Common Mistakes

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

Why This Works: The Logic of "Exactly One"

Step-by-Step

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

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