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Two dice are tossed together. Find the Probability of getting a doublet or total of 6 |
Answer» Let ‘S’ be the sample space. So, n(S) = 62 = 36. Also, Let ‘A’ and ‘B’ be the events of getting a doublet and a total of 6 respectively. ∴ A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} ⇒ n(A) = 6, n(B) = 5 B = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) ∴ p(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{6}{36}\) , P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{5}{36}\) A ∩ B = {(3, 3)} ⇒ n(A ∩ B) = 1 ⇒ P(A ∩ B) = \(\frac{n(A∩B)}{n(S)}\) = \(\frac{1}{36}\) We know– P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = \(\frac{6}{36}\) + \(\frac{5}{36}\) − \(\frac{1}{36}\) = \(\frac{10}{36}\) = \(\frac{5}{18}\) |
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