is it c.

Explanation:

In the xy-coordinate plane, a rectangle has the vertices (3.3, 3.3), (3.3, -7.3), (-2.3, -7.3), (-2.3, 3.3).
P is the probability of picking a point (x, y) randomly from within the rectangle, such that x and y are both positive integers.

Let us plot the points on xy-plane (in rough)
Notice that \(x\) (integer values) ranges from -2 to 3 i.e. 6 values (-2, -1, 0, 1, 2, and 3)
and \(x\) (integer values) ranges from -7 to 3 i.e. 11 values (-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, and 3)

So, total number of possibilities of picking a point with both the coordinates as integer = (6)(11) = 66

(Refer to the graph)
The favorable outcomes are the points in the Ist quadrant i.e. 9

Col. A: \(P = \frac{9}{66} = \frac{3}{22}\)
Col. B: \(\frac{1089}{5936}\)

Col. A: 0.136
Col. B: 0.183

Hence, option B

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??

Explanation:

In the xy-coordinate plane, a rectangle has the vertices (3.3, 3.3), (3.3, -7.3), (-2.3, -7.3), (-2.3, 3.3).
P is the probability of picking a point (x, y) randomly from within the rectangle, such that x and y are both positive integers.

Let us plot the points on xy-plane (in rough)
Notice that \(x\) (integer values) ranges from -2 to 3 i.e. 6 values (-2, -1, 0, 1, 2, and 3)
and \(x\) (integer values) ranges from -7 to 3 i.e. 11 values (-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, and 3)

So, total number of possibilities of picking a point with both the coordinates as integer = (6)(11) = 66

(Refer to the graph)
The favorable outcomes are the points in the Ist quadrant i.e. 9

Col. A: \(P = \frac{9}{66} = \frac{3}{22}\)
Col. B: \(\frac{1089}{5936}\)

Col. A: 0.136
Col. B: 0.183

Hence, option B

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?