Correct Answer - Option 3 :
\(P(A|B)\ge\dfrac{L+M-1}{M}\)
Concept:
\(\rm P(A|B)=\dfrac{P(A∩ B)}{P(B)}\)
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Calculation:
Given P(A) = L and P(B) = M
Here P(A ∪ B) ≤ 1 (∵ Max value of probability is 1)
P(A) + P(B) - P(A ∩ B) ≤ 1
L + M - P(A ∩ B) ≤ 1
P(A ∩ B) ≥ L + M - 1
Now \(\rm P(A|B)=\dfrac{P(A∩ B)}{P(B)}\)
\(\rm P(A|B)\geq\dfrac{L+M-1}{M}\)