Suppose Y is distributed uniformly in the open interval (1,6). The probability that the polynomial 3x² + 6xY + 3Y + 6 has only real roots is (rounded to 1 decimal place) _________

Answer: 0.8Concept: Discriminant of a Quadratic and ProbabilityFor a quadratic ax² + bx + c, real roots exist iff discriminant D = b² - 4ac ≥ 0. Here: a=3, b=6Y, c=3Y+6. D = (6Y)² - 4(3)(3Y+6) = 36Y² - 12(3Y+6) = 36Y² - 36Y - 72 = 36(Y² - Y - 2) = 36(Y-2)(Y+1).D ≥ 0 iff (Y-2)(Y+1) ≥ 0. Since Y ∈ (1,6), Y+1 > 0 always. So condition reduces to Y-2 ≥ 0, i.e., Y ≥ 2.P(Y ≥ 2) = P(2 ≤ Y Formula: For a uniform distribution on (a,b), P(c ≤ Y ≤ d) = (d-c)/(b-a). Discriminant condition: D ≥ 0 ensures real roots.

GATE-CS is a competitive examination conducted in India. This question is from the 2019 GATE-CS Engineering Mathematics paper.

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