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Three randomly chosen nonnegative integers x, y and z are found to satisfy the equation x + y + z = 10. Then the probability that z is even, is(A) 36/55 (B) 6/11 (C) 1/2 (D) 5/11 |
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Answer» Correct option is (B) 6/11 Total number of solutions = 10 + 3 – 1C3 – 1 = 66 Favourable number of solutions = 11C1 + 9C1 + 7C1 + 5C1 + 3C1 + 1C1 = 36 P(req) = \(\frac{36}{66}=\frac{6}{11}\) |
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