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An unbiased die is tossed twice. What is the probability of getting a 4, 5 or 6 on the first toss and a 1, 2, 3 or 4 on the second toss?A. \(\frac{1}{3}\)B. \(\frac{2}{3}\)C. \(\frac{3}{4}\)D. \(\frac{5}{6}\) |
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Answer» Correct answer is A. A die is tossed twice, The probability of getting a 4, 5 or 6 in the first trial is 3/6 = P(A) The probability of getting a 1, 2, 3 or 4 in the second trial is 4/6 =P(B) As the events are independent, the probability of these two events together will be, P(A).P(B) = 1/3. |
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