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NCERT Exemplar · Q35

Q.If P(A)=0.4P(A) = 0.4, P(B)=0.8P(B) = 0.8 and P(B∣A)=0.6P(B \mid A) = 0.6, then P(A∪B)P(A \cup B) is equal to
(A) 0.240.24
(B) 0.30.3
(C) 0.480.48
(D) 0.960.96

Rajasthan RbseMCQ· 1mImportance★★★★★
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The key is to first find P(A∩B)P(A \cap B) using the conditional probability formula P(B∣A)=P(A∩B)P(A)P(B \mid A) = \frac{P(A \cap B)}{P(A)}, then apply the addition rule P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). The result is 0.960.96.

We are given P(A)=0.4P(A) = 0.4, P(B)=0.8P(B) = 0.8, and P(B∣A)=0.6P(B \mid A) = 0.6. The goal is P(A∪B)P(A \cup B).

The core idea: conditional probability tells us how two events overlap. Once we know the overlap (A∩BA \cap B), the union follows directly from the addition rule. Without the overlap, we cannot simply add P(A)P(A) and P(B)P(B) — that would double-count the intersection.

  1. Find P(A∩B)P(A \cap B) using the definition of conditional probability.

    By definition, P(B∣A)=P(A∩B)P(A)P(B \mid A) = \frac{P(A \cap B)}{P(A)}.

    Rearranging: P(A∩B)=P(B∣A)⋅P(A)P(A \cap B) = P(B \mid A) \cdot P(A).

    Substitute: P(A∩B)=0.6×0.4=0.24P(A \cap B) = 0.6 \times 0.4 = 0.24.

    Watch out

    A common mistake is to think P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) — that is only true if AA and BB are independent. Here, P(B∣A)=0.6P(B \mid A) = 0.6 while P(B)=0.8P(B) = 0.8, so they are not independent. Always use the conditional formula when given P(B∣A)P(B \mid A).

  2. Apply the addition rule for probability. …

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