Question

Choose a random real number uniformly from the unit interval Ω = [0, 1]. Consider the events: A = [1/2,1], B = [1/2,3/4], C = [1/16,9/16]. Show that P(A ∩ B ∩ C) = P(A) · P(B) · P(C), but the events A, B, C are not mutually independent.

Answer 1

Let x be a random real number belonging to the unit interval [0,1].

A = [1/2,1]

B = [1/2,3/4]

C = [1/16,9/16]

A ∩ B = [1/2,3/4]

(A ∩ B ∩ C) = [1/2, 9/16]

P(A) = 1 - 1/2 = 1/2

P(B) = 3/4 - 1/2 = 1/4

P(C) = 9/16 - 1/16 = 1/2

P(A ∩ B ∩ C) = 9/16 - 1/2 = 1/16

P(A) · P(B) · P(C) = 1/2 . 1/4 . 1/2 = 1/16

It seems so that events A, B and C are independent. Let's check about the independence of events taken in pairs. Mutual independence means that all the events are independent irrespective of the number of the events under consideration.

P(A ∩ B) = 3/4 - 1/2 = 1/4

P(A).P(B) = 1/2 . 1/4 = 1/8 { NOT EQUAL TO P(A ∩ B) }

In the same way we can show that A & C are not independent and B & C are not independent.

Thanks.