Addition Theorem of Events MCQ Quiz in मल्याळम - Objective Question with Answer for Addition Theorem of Events - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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