If A and B are two independent events such that P ( A ∩ B ) = 1 6 and P ( A ¯ ∩ B ¯ ) = 1 3 , then write the values of P ( A ) and P ( B ) .

Given, P ( A ∩ B ) = 1 6 and P ( A ¯ ∩ B ¯ ) = 1 3 By De Morgan's law, P ( A ¯ ∩ B ¯ ) = P ( A ∪ B ¯ ) We know P ( A ¯ ∩ B ¯ ) = 1 - P ( A ∪ B ) ⇒ P ( A ∪ B ) = 1 - P ( A ¯ ∩ B ¯ ) = 1 - 1 3 = 2 3 From the addition theorem of probability, we know: P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) ∴ P ( A ) + P ( B ) = P ( A ∪ B ) + P ( A ∩ B ) ⇒ P ( A ) + P ( B ) = 2 3 + 1 6 = 4 + 1 6 = 5 6 Thus, P ( A ) + P ( B ) = 5 6 .......(i) A and B are independent events. Again, P ( A ∩ B ) = P ( A ) × P ( B ) = 1 6 We know that, a - b 2 = a + b 2 - 4 a b By formula, we have: { P ( A ) - P ( B ) } 2 = { P ( A ) + P ( B ) } 2 - 4 × P ( A ) × P ( B ) ⇒ { P ( A ) - P ( B ) } 2 = 5 6 2 - 4 6 = 25 36 - 4 6 = 25 - 24 36 = 1 36 ∴ P ( A ) - P ( B ) = 1 6 .......(ii) Adding eq.(i) and eq.(ii), 2 P A = 5 6 + 1 6 ⇒ 2 P ( A ) = 1 ⇒ P A = 1 2 From (i) eq., 1 2 + P B = 5 6 ⇒ P B = 5 6 - 1 2 = 5 - 3 6 = 2 6 = 1 3 Hence, P ( A ) = 1 2 and P ( B ) = 1 3 .