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 \(\rm P(A|B)=\dfrac{P(A∩ B)}{P(B)}\)

\(\rm P(A|B)\geq\dfrac{L+M-1}{M}\)