The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is

The correct option is 0.25.

To find the probability that the sum of the variables $x+y$ is greater than 20, we can analyze the sample space geometrically.

The variable $x$ is uniformly distributed between 0 and 10, and the variable $y$ is uniformly distributed between 0 and 20. Since $x$ and $y$ are independent, the joint sample space is a rectangle in the $xy$-plane defined by:
$0\le x\le 10$ and $0\le y\le 20$.

The total area of this rectangular sample space is:
$\text{Total Area}=10\times 20=200$.

We want to find the probability that:
$x+y>20$, which can be rewritten as $y>20-x$.

Let us identify the region within our rectangle where this inequality holds. The boundary line is:
$y=20-x$.
Let's find the intersection points of this line with the boundaries of the rectangular region:
- When $x=0$, $y=20$. This is the top-left corner of the rectangle, $(0,20)$.
- When $x=10$, $y=10$. This point lies on the right boundary of the rectangle, $(10,10)$.

The region where $y>20-x$ within the rectangle is a right-angled triangle at the top-right corner. The vertices of this triangular region are:
$(0,20)$, $(10,20)$, and $(10,10)$.

The base of this triangle along the top edge ($y=20$) goes from $x=0$ to $x=10$, so its length is:
$\text{Base}=10-0=10$.
The height of this triangle along the right edge ($x=10$) goes from $y=10$ to $y=20$, so its length is:
$\text{Height}=20-10=10$.

The area of this triangular region is:
$\text{Favorable Area}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 10\times 10=50$.

Since the probability distribution is uniform, the probability is the ratio of the favorable area to the total area:
$P(x+y>20)=\frac{\text{Favorable Area}}{\text{Total Area}}=\frac{50}{200}=0.25$.