Addition Theorem of Events MCQ Quiz in தமிழ் - Objective Question with Answer for Addition Theorem of Events - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Mar 23, 2025
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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