Question

Let 0 < P (1) < 1, 0 < P (2) < 1 and P (A È B) = P (1) + P (2)– P (1) P (2), then -

• A. P (A | B) = 0
• B. P (B | A) = 0
• C. P (A¢∩ B¢) = P (A¢) P (B¢)
• D. P (A | B) + P (B | A) = 1

Answer 1

Answer: (C)

P (A¢∩ B¢) = P (A¢) P (B¢)

Answer 2

We know that

P (A È B) = P (1) + P (2) – P (A Ç B)

̃ P (1) + P (2) – P (1) P (2) = P (1) + P (2) – P (A Ç B)

̃ P (1) . P (2) = P (A Ç B) ̃ A and B are independent.

Thus, P (A¢ Ç B¢) = P (A¢) P (B¢)