Probability
Probability
star_batch_jee_advanced_2025
Grade None
Question:
If $A$ and $B$ are exhaustive events in a sample space such that probabilities of the events $A \cap B$, $A$, $B$ and $A \cup B$ are in A.P. If $P(A) = K$, where $0 < K \leq 1$, then:
$P(B) = \frac{K+1}{2}$
$P(A \cap B) = \frac{3K-1}{2}$
$P(A \cup B) = 1$
$P(A' \cup B') = \frac{3(1-K)}{2}$
Step-by-Step Solution
Key Concept: Exhaustive events mean their union covers the entire sample space, so $P(A ∪ B) = 1$, which constrains all probability relationships.
Since $A$ and $B$ are exhaustive events, $A ∪ B = S$ and $P(A ∪ B) = 1$. Using $2P(B) = P(A) = P(A ∪ B) = K+1$, we get $P(B) = \frac{K+1}{2}$ and $P(A ∩ B) = \frac{3K-1}{2}$. Then $P(A' ∪ B') = 1 - P(A ∩ B) = \frac{3(1-K)}{2}$.
Correct Answer: 1,2,3,4