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Suppose $Y$ is distributed uniformly in the open interval $(1,6)$. The probability that the polynomial $3x^2 +6xY+3Y+6$ has only real roots is (rounded off to $1$ decimal place) _______

Here, $Y$ is uniformly distributed in $(1,6)$

$\therefore \: f(Y) = \begin{cases} \frac{1}{b-a} & 1<Y<6 \\ 0, & \text{otherwise} \end{cases}$

$\implies f(Y)=\frac{1}{b-a} = \frac{1}{6-1} = \frac{1}{5}$

Now given Polynomial,

$3x^2+6xY+(3Y+6)$

For real roots $b^2-4ac \geq 0$

$\implies (6Y)^2 - 4 \times 3 \times (3Y+6) \geq 0$

$\implies Y^2-2Y+Y-2 \geq 0$

$\implies (Y-2)(Y+1) \geq 0$

$\implies Y \geq 2, \: Y \geq -1$

So, we need area of the region in the above graph, for $(2,6)$

Area $= B \times H$

$=(6-2) \times \frac{1}{5}$

$= \frac{4}{5} = 0.8$

The distribution is continuous by default, so the open / closed interval should be irrelevant.
This 2 mark question tells us how important the formula for real roots is. We neglect it. And yes it is asked in 2 marks. I didn't attempt it

(Y-2)(Y+1) >= 0

It is solved wrongly in best answer.

Y >= 2     and    Y <= -1

For real roots, b^2 - 4ac >= 0

(6Y)^2 - 4*3*(3Y+6)>=0

36Y^2 - 36Y - 72 >=0

36(Y+1)(Y-2)>=0, This equation is 0 at Y = -1 and 2. We need to consider for range (1,6) so from (2,6) it is >= 0. You can plot graph to observe this or simple value substitution above 2 will do.

Now probability = (1/(6-1)) * (6 - 2) = (1/5)* 4 = 0.8

ok, so we are taking a part of open interval.

Why donot r we taking full total interval between $\left ( 1,6 \right )$

This range is already given in question.

right?

https://gateoverflow.in/302801/gate2019-47?show=305183#c305183

No, I want to know, why have not the interval between (1,2) in the probability calculation. What is reason behind it?

Y is distributed in open interval (1,6) means value of y can be from 2,3,4 or 5. And for real roots b^2-4ac>=0. For all y values this condition is satisfied. so probability should be 1. so, my answer is 1. Correct me if i am wrong.