Addition Theorem of Events MCQ Quiz in தமிழ் - Objective Question with Answer for Addition Theorem of Events - இலவச PDF ஐப் பதிவிறக்கவும்

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