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Let us consider two independent events A and B, then P(A ∩ B) = P(A). P(B)

when an unbiased die is thrown twice

S ={(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)

(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}

Let us describe two events as

A: odd number on the first throw

B: odd number on the second throw

To find P(A)

A = {(1, 1), (1, 2), (1, 3), …, (1, 6)

(3, 1), (3, 2), (3, 3), …, (3, 6)

(5, 1), (5, 2), (5, 3), …, (5, 6)}

Thus, P (A) = 18/36 = 1/2

To find P(B)

B = {(1, 1), (2, 1), (3, 1), …, (6, 1)

(1, 3), (2, 3), (3, 3), …, (6, 3)

(1, 5), (2, 5), (3, 5), …, (6, 5)}

Thus, P (B) = 18/36 = 1/2

A ∩ B = odd number on the first & second throw = { (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}

So, P(A ∩ B) = 9/36 = 1/ 4

Now, P(A). P(B) = (1/2) × (1/2) = 1/4

As P(A ∩ B) = P(A). P(B),

Hence, the two events A and B are independent events.

Regards