The probabilities of events $A \cap B$, $A$, $B$ and $A \cup B$ are respectively in A.P. with second term equal to the common difference. Therefore, $A$ and $B$ are:
Step-by-Step Solution
Key Concept: Events A and B are in A.P. with P(A) = d as the common difference, so P(A∩B) = 0 (mutual exclusivity). Use the constraint P(A∪B) = P(A) + P(B) - P(A∩B) along with the A.P. condition a + d = d to derive a = 0, then verify P(B) = 2P(A) establishes the second relationship.
With the given conditions $P(A ∩ B) = a$, $P(A) = a + d$, $P(B) = a + 2d$, and $P(A ∪ B) = a + 3d$, plus the constraint $a + d = d$, we derive $a = 0$. This yields $P(A ∩ B) = 0$, $P(A) = d$, $P(B) = 2d$, and $P(A ∪ B) = 3d$.
Correct Answer: 1,4