Class CBSE Class 12 Mathematics Probability Q #1509
COMPETENCY BASED
APPLY
3 Marks 2026 AISSCE(Board Exam) SA
The probability of simultaneous occurrence of at least one of the two events $X$ and$Y$ is $a$. If the probability that exactly one of the events $X, Y$ occurs is $b$, prove that $P(X') + P(Y') = 2 – 2a + b$.
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Detailed Solution

Step 1: Define the given probabilities

Let $P(X) = x$ and $P(Y) = y$. The probability of at least one event occurring is $P(X \cup Y) = a$. The probability of exactly one event occurring is $P(X \cup Y) - P(X \cap Y) = b$.

Step 2: Express the intersection

Since $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = a$, we have $P(X \cap Y) = P(X) + P(Y) - a$. Substituting this into the expression for $b$: $$b = P(X \cup Y) - P(X \cap Y) = a - (P(X) + P(Y) - a) = 2a - (P(X) + P(Y))$$ Thus, $P(X) + P(Y) = 2a - b$.

Step 3: Evaluate the target expression

We need to find $P(X') + P(Y')$. Using the complement rule $P(E') = 1 - P(E)$: $$P(X') + P(Y') = (1 - P(X)) + (1 - P(Y)) = 2 - (P(X) + P(Y))$$

Step 4: Final Substitution

Substitute the value of $P(X) + P(Y)$ derived in Step 2: $$P(X') + P(Y') = 2 - (2a - b) = 2 - 2a + b$$ This completes the proof.

Final Answer: Proved: P(X') + P(Y') = 2 - 2a + b

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to translate verbal probability conditions into algebraic equations and manipulate them to reach a proof.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a specific sequence of algebraic steps involving set theory and probability axioms to derive the identity.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This question tests the conceptual understanding of the Addition Theorem of Probability and the relationship between events and their complements, which is a core competency in the Probability chapter.

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