Question:
If A and B are two events such that \(P(A) = \dfrac{1}{2}\), \(P(B) = \dfrac{1}{3}\), \(P\left(\dfrac{A}{B}\right) = \dfrac{1}{4}\) then \(P(A’ \cap B’)\) equals
यदि A और B दो ऐसी घटनाएँ हैं कि \(P(A) = \dfrac{1}{2}\), \(P(B) = \dfrac{1}{3}\), \(P\left(\dfrac{A}{B}\right) = \dfrac{1}{4}\) है, तो \(P(A’ \cap B’)\) किसके बराबर है?
Options:
- \(\dfrac{1}{12}\)
- \(\dfrac{3}{16}\)
- \(\dfrac{1}{4}\)
- \(\dfrac{3}{4}\)
- \(\dfrac{1}{12}\)
- \(\dfrac{3}{16}\)
- \(\dfrac{1}{4}\)
- \(\dfrac{3}{4}\)
Correct Answer: c)
Explanation:
\(P(A \cap B) = P(A/B) \cdot P(B) = \dfrac{1}{4} \times \dfrac{1}{3} = \dfrac{1}{12}\).
\(P(A \cup B) = P(A) + P(B) – P(A \cap B) = \dfrac{1}{2} + \dfrac{1}{3} – \dfrac{1}{12} = \dfrac{9}{12} = \dfrac{3}{4}\).
\(P(A’ \cap B’) = 1 – P(A \cup B) = 1 – \dfrac{3}{4} = \dfrac{1}{4}\).
\(P(A \cap B) = \dfrac{1}{4} \times \dfrac{1}{3} = \dfrac{1}{12}\)।
\(P(A \cup B) = \dfrac{1}{2} + \dfrac{1}{3} – \dfrac{1}{12} = \dfrac{3}{4}\)।
\(P(A’ \cap B’) = 1 – P(A \cup B) = 1 – \dfrac{3}{4} = \dfrac{1}{4}\)।