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Re: In the xy-coordinate plane, a rectangle has the vertices (3.3, 3.3), [#permalink]
void wrote:
is it c.


Please provide a detailed solution.
The OA will be revealed soon
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Re: In the xy-coordinate plane, a rectangle has the vertices (3.3, 3.3), [#permalink]
KarunMendiratta wrote:
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.
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Re: In the xy-coordinate plane, a rectangle has the vertices (3.3, 3.3), [#permalink]
kapil1 wrote:
KarunMendiratta wrote:
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??
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Re: In the xy-coordinate plane, a rectangle has the vertices (3.3, 3.3), [#permalink]
1
KarunMendiratta wrote:
kapil1 wrote:
KarunMendiratta wrote:
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??


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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In the xy-coordinate plane, a rectangle has the vertices (3.3, 3.3), [#permalink]
1
KarunMendiratta wrote:
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 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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Re: In the xy-coordinate plane, a rectangle has the vertices (3.3, 3.3), [#permalink]
kapil1 wrote:

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


I guess I forgot to mention the integers in question.
Thanks for pointing out my mistake.
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Re: In the xy-coordinate plane, a rectangle has the vertices (3.3, 3.3), [#permalink]
1
COolguy101 wrote:
KarunMendiratta wrote:
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 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?


Yes dear, made the necessary changes.
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