retagged by
16,078 views

4 Answers

Best answer
68 votes
68 votes

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 $= B \times H$

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

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

edited by
11 votes
11 votes
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
1 votes
1 votes

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.

Answer:

Related questions

24 votes
24 votes
2 answers
2
Arjun asked Feb 12, 2020
12,125 views
For $n>2$, let $a \in \{0,1\}^n$ be a non-zero vector. Suppose that $x$ is chosen uniformly at random from $\{0,1\}^n$. Then, the probability that $\displaystyle{} \Sigm...
26 votes
26 votes
4 answers
3
Arjun asked Feb 7, 2019
19,276 views
Consider the following matrix:$R = \begin{bmatrix} 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \\ 1 & 4 & 16 & 64 \\ 1 & 5 & 25 & 125 \end{bmatrix}$The absolute value of the product ...