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An fair die is thrown double times. Assume that the event A is “odd number on the first throw” and B the event “odd number on the second throw”. Compare the independence of the events A and B. An fair die is thrown double times. Assume that the event A is “odd number on the first throw” and B the event “odd number on the second throw”. Compare the independence of the events A and B.
Welcome to AskiitiansLet us consider two independent events A and B, then P(A ∩ B) = P(A). P(B)when an unbiased die is thrown twiceS ={(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 asA: odd number on the first throwB: odd number on the second throwTo 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 = 1/2To 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 = 1/2A ∩ 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/ 4Now, P(A). P(B) = (1/2) × (1/2) = 1/4As P(A ∩ B) = P(A). P(B),Hence, the two events A and B are independent events.Regards
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