GATE CS 2019 Q47 — Probability: Random Variables and Distributions

For a quadratic equation $Ax^2 + Bx + C = 0$ to have only real roots, its discriminant $D = B^2 - 4AC$ must be greater than or equal to zero ($D \ge 0$).

Given the polynomial:
$$3x^2 + 6Yx + (3Y + 6)$$

Here, the coefficients are:

• $A = 3$
• $B = 6Y$
• $C = 3Y + 6$

Setting the discriminant $D \ge 0$: $$(6Y)^2 - 4(3)(3Y + 6) \ge 0$$ $$36Y^2 - 12(3Y + 6) \ge 0$$

Dividing the entire inequality by $12$: $$3Y^2 - (3Y + 6) \ge 0$$ $$3Y^2 - 3Y - 6 \ge 0$$

Dividing by $3$: $$Y^2 - Y - 2 \ge 0$$ $$(Y - 2)(Y + 1) \ge 0$$

This inequality holds when: $$Y \le -1 \quad \text{or} \quad Y \ge 2$$

Since $Y$ is uniformly distributed in the open interval $(1, 6)$, the probability density function of $Y$ is: $$f(y) = \frac{1}{6 - 1} = \frac{1}{5} \quad \text{for } 1 < y < 6$$

The condition $Y \le -1$ is outside the interval $(1, 6)$ and has a probability of $0$. Thus, we only need to find the probability that $Y \ge 2$ within the interval $(1, 6)$: $$P(Y \ge 2) = \int_{2}^{6} \frac{1}{5} \, dy = \frac{6 - 2}{5} = \frac{4}{5} = 0.8$$