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 -2.3 to 3.3 i.e. 5.6
and \(x\) (integer values) ranges from -7.3 to 3.3. i.e. 10.6

So, total number of possibilities of picking a point will be the area of the rectangle = (5.6)(10.6) = 59.36

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

Col. A: \(P = \frac{9}{59.36} = \frac{900}{5936}\)
Col. B: \(\frac{1089}{5936}\)

Hence, option B

Edited on 13-Aug 2021

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is it c.

Please provide a detailed solution.
The OA will be revealed soon
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Please see the denominator in Quantity A will be 5936 only as your are picking a point (x,y) from within the rectangular region.
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Yes, we are picking a point with 2 coordinates. Can you please explain how will we get 5936 as the denominator??
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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
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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?
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I guess I forgot to mention the integers in question.
Thanks for pointing out my mistake.
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Yes dear, made the necessary changes.
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