Please provide a detailed solution.
The OA will be revealed soon

Please see the denominator in Quantity A will be 5936 only as your are picking a point (x,y) from within the rectangular region.

Yes, we are picking a point with 2 coordinates. Can you please explain how will we get 5936 as the denominator??

Total number of outcome of picking a point (x, y) randomly from within the rectangle = Area of the given rectangle = 10.6 x 5.6 = 59.36
Total number of desired outcome such that x and y are both positive integers = Area of the square in the first quadrant = 3 x 3 = 9

Hence, probability of picking a point (x, y) randomly from within the rectangle, such that x and y are both positive integers = 9/59.36 = 900/5936

Total no of possibility must be area of that rectangle, is not it sir?, Why do we need to consider that picking point must be integer for total probability?

I guess I forgot to mention the integers in question.
Thanks for pointing out my mistake.

Yes dear, made the necessary changes.

@karunmehndiratta the area under the region of square will be 3.3*3.3= 10.89

answer will be C

correct answer is C not B

Given the rectangle with vertices:

$$
\((3.3,3.3),(3.3,-7.3),(-2.3,-7.3),(-2.3,3.3),\)
$$


Step 1: Calculate the side lengths
- Length along the $x$-axis:

$$
\(|3.3-(-2.3)|=3.3+2.3=5.6 \text {. }\)
$$

- Length along the $y$-axis:

$$
\(|3.3-(-7.3)|=3.3+7.3=10.6\)
$$


Step 2: Calculate area of the rectangle

$$
\(\text { Area }=5.6 \times 10.6=59.36\)
$$


Step 3: Count the positive integer points ( $x, y$ ) inside the rectangle
- Possible $x$ integer values:

Since $x$ ranges from -2.3 to 3.3 , the positive integers satisfying this are:

$$
\(1,2,3,\)
$$

so 3 values.
- Possible $y$ integer values:

Since $y$ ranges from -7.3 to 3.3 , positive integers satisfying this are:

$$
\(1,2,3,\)
$$

so 3 values.

Step 4: Number of points with positive integer $x$ and $y$

$$
\(3 \times 3=9\) .
$$

Step 5: Calculate probability $P$ :

$$
\(P=\frac{\text { Number of positive integer points }}{\text { Area of rectangle }}=\frac{9}{59.36} \approx 0.1516\) .
$$


Step 6: Compare with Quantity B:

$$
\(\frac{1089}{5936} \approx \frac{9^2}{(59.36 \times 100)} \approx 0.1833\)
$$

which is greater than 0.1516 .
Final comparison:
- Quantity A $\(=P \approx 0.1516\)$
- Quantity $\(B=\frac{1089}{5936} \approx 0.1833\)$

So:
Quantity A $\(<\)$ Quantity B.
Answer: Quantity B is greater.