Addition Theorem of Events MCQ Quiz in मल्याळम - Objective Question with Answer for Addition Theorem of Events - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 12, 2025
Latest Addition Theorem of Events MCQ Objective Questions
Top Addition Theorem of Events MCQ Objective Questions
Addition Theorem of Events Question 1:
Let A and B are two independent events. The probability that both A and B happen is \(\frac{1}{12}\) and probability that neither A nor B happen is \(\frac{1}{2}\) Then
Answer (Detailed Solution Below)
Addition Theorem of Events Question 1 Detailed Solution
Calculation
P(A∩B) = \(\frac{1}{12}\) ⇒ P(A)P(B) = \(\frac{1}{12}\) ...(1)
P(A'∩B') = \(\frac{1}{2}\) ⇒ P(A∪B) = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\)
⇒ P(A) + P(B) - P(A)P(B) = \(\frac{1}{2}\)
⇒ P(A) + P(B) = \(\frac{1}{2}\) + \(\frac{1}{12}\) = \(\frac{7}{12}\)
From (1),
P(A)(\(\frac{7}{12}\) - P(A)) = \(\frac{1}{12}\)
⇒ P(A)² - (\(\frac{7}{12}\))P(A) + \(\frac{1}{12}\) = 0
⇒ 12P(A)² - 7P(A) + 1 = 0
⇒ 12P(A)² - 4P(A) - 3P(A) + 1 = 0
⇒ (3P(A) - 1)(4P(A) - 1) = 0
∴ P(A) = \(\frac{1}{3}\) or P(A) = \(\frac{1}{4}\)
P(B) = \(\frac{1}{12}\) × 3 ⇒ P(B) = \(\frac{1}{4}\)
P(B) = \(\frac{1}{12}\) × 4 ⇒ P(B) = \(\frac{1}{3}\)
∴ P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{4}\)
Hence option 1 is correct