Suppose 𝑌 is distributed uniformly in the open interval (1,6). The…
2019
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 off to 1 decimal place) ________.
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Correct answer: 0.8
Key idea: use the discriminant of the quadratic in x to find for which values of Y the roots are real.
Identify coefficients: the quadratic is 3x² + 6xY + (3Y + 6), so a = 3, b = 6Y, c = 3Y + 6.
Compute the discriminant: D = b² - 4ac = (6Y)² - 4·3·(3Y + 6) = 36Y² - 12(3Y + 6) = 36Y² - 36Y - 72 = 36(Y² - Y - 2) = 36(Y - 2)(Y + 1).
Require D ≥ 0 for real roots, so (Y - 2)(Y + 1) ≥ 0, which gives Y ≤ -1 or Y ≥ 2. Given Y is uniform on (1, 6), the relevant values are Y ≥ 2.
Measure the favorable set: within (1, 6) the favorable interval is [2, 6), whose length is 4. The total length of (1, 6) is 5. Thus the probability = 4/5 = 0.8.
Answer (rounded to 1 decimal place): 0.8