Question 52 of 55intermediate🔧 ApplyShort Answer4 marks

A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (i) P(A ∪ B) (ii) P(A′ ∩ B′) (iii) P(A ∩ B′) (iv) P(B ∩ A′)

(i)
Answer

P(A ∪ B): Using the addition theorem of probability found in the text, P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). Substituting the given values: P(A∪B)=0.54+0.69−0.35=0.88P(A \cup B) = 0.54 + 0.69 - 0.35 = 0.88.

(ii)
Answer

P(A′ ∩ B′): This represents the probability of neither A nor B occurring. Using the relation derived from set theory and complement probability, P(A′∩B′)=P((A∪B)′)=1−P(A∪B)P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B). Substituting the value from (i): 1−0.88=0.121 - 0.88 = 0.12.

(iii)
Answer

P(A ∩ B′): This represents the probability that A occurs but B does not. Using the partition of A, P(A∩B′)=P(A)−P(A∩B)P(A \cap B') = P(A) - P(A \cap B). Substituting values: 0.54−0.35=0.190.54 - 0.35 = 0.19.

(iv)
Answer

P(B ∩ A′): This represents the probability that B occurs but A does not. Using the partition of B, P(B∩A′)=P(B)−P(A∩B)P(B \cap A') = P(B) - P(A \cap B). Substituting values: 0.69−0.35=0.340.69 - 0.35 = 0.34.

Explanation

The solution is derived using the fundamental probability theorems present in the provided context. Part (i) uses the Addition Theorem explicitly shown in the derivation for three events in the text (P(A∪E)=P(A)+P(E)−P(A∩E)P(A \cup E) = P(A) + P(E) - P(A \cap E)). Parts (ii), (iii), and (iv) utilize set-theoretic properties implied by the questions in the context (like Q15 and Q17) regarding complements and disjoint partitions of events.

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