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Point P(a,b) is randomly selected in the region enclosed by

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Point P(a,b) is randomly selected in the region enclosed by  [#permalink]

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New post Updated on: 27 Sep 2013, 01:35
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Point P(a,b) is randomly selected in the region enclosed by the following lines: 3y + 2x = 6, x=0 and y = 0. What is the probability that b>a ?

I don't have OA for this question...

Can someone please explain me how to calculate the probability?
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Originally posted by jellybean23 on 26 Sep 2013, 23:51.
Last edited by Bunuel on 27 Sep 2013, 01:35, edited 1 time in total.
Edited the question.
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Re: Point P(a,b) is randomly selected in the region enclosed by  [#permalink]

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New post 27 Sep 2013, 01:57
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KK23 wrote:
Point P(a,b) is randomly selected in the region enclosed by the following lines: 3y + 2x = 6, x=0 and y = 0. What is the probability that b>a ?

I don't have OA for this question...

Can someone please explain me how to calculate the probability?


Refer to the diagram below:
Attachment:
Probability.png
Probability.png [ 13.38 KiB | Viewed 9431 times ]

The area of the triangle enclosed by 3y + 2x = 6 (y=2-2/3*x), x=0 and y = 0 is 1/2*2*3=3.

We need the y-coordinate of P to be greater than its x-coordinate (y>x), so point P must be above line y=x and in the enclosed region.

The probability of that is (favorable)/(total)=(yellow triangle)/(entire triangle).

Now, we need to find intersection point of y=x and 3y+2x=6, which is (6/5, 6/5). Thus the area of yellow triangle is 1/2*6/5*2=6/5.

P=(6/5)/3=2/5.

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in-the-coordinate-plane-rectangular-region-r-has-vertices-a-104869.html
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Hope it helps.
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Re: Point P(a,b) is randomly selected in the region enclosed by  [#permalink]

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New post 20 Jul 2014, 10:12
Hi Bunuel,
Can you explain how you calculated the area of the triangle?
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Re: Point P(a,b) is randomly selected in the region enclosed by  [#permalink]

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New post 20 Jul 2014, 11:33
1
ronr34 wrote:
Hi Bunuel,
Can you explain how you calculated the area of the triangle?


Look at the diagram below:
Attachment:
Untitled.png
Untitled.png [ 13.67 KiB | Viewed 8741 times ]
The base of the triangle is 2 (vertical green line) and the height of the triangle is 6/5 (horizontal green line, x-coordinate of the intersection point). So, the area is /2*6/5*2=6/5.

Check other Probability and Geometry questions in our Special Questions Directory.
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Re: Point P(a,b) is randomly selected in the region enclosed by  [#permalink]

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New post 31 Jul 2016, 04:10
Bunuel wrote:
Point P(a,b) is randomly selected in the region enclosed by the following lines: 3y + 2x = 6, x=0 and y = 0. What is the probability that b>a ?

I don't have OA for this question...

Can someone please explain me how to calculate the probability?

Refer to the diagram below:
Attachment:
Probability.png

The area of the triangle enclosed by 3y + 2x = 6 (y=2-2/3*x), x=0 and y = 0 is 1/2*2*3=3.

We need the y-coordinate of P to be greater than its x-coordinate (y>x), so point P must be above line y=x and in the enclosed region.

The probability of that is (favorable)/(total)=(yellow triangle)/(entire triangle).

Now, we need to find intersection point of y=x and 3y+2x=6, which is (6/5, 6/5). Thus the area of yellow triangle is 1/2*6/5*2=6/5.

P=(6/5)/3=2/5.

Similar questions to practice:
set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html
in-the-coordinate-plane-rectangular-region-r-has-vertices-a-104869.html
in-the-xy-plane-a-triangle-has-vertexes-0-0-4-0-and-86926.html
in-the-xy-plane-a-triangle-has-vertexes-0-0-4-0-and-88395.html

Hope it helps.


Quote:
Bunuel Sir, two questions here,

Why are you considering "y=x" line.
How did you get 6/5?

Please elaborate.
Thanks in advance.
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Re: Point P(a,b) is randomly selected in the region enclosed by  [#permalink]

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New post 27 Jan 2017, 02:27
how do we get this point of intersection?
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Re: Point P(a,b) is randomly selected in the region enclosed by  [#permalink]

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New post 23 Feb 2017, 22:52
GovindAgrawal wrote:
how do we get this point of intersection?


Hi,

Let me try explaining. It took me a while figuring it out myself.

First, we can replace the lines x=0 and y=0 by a line x=y. That leaves us with just 2 lines :- 1st is x=y and 2nd is 3y= -2x + 6.

To find the intersection of these 2 lines, equate the RHS variables and you'll get x=6/5. [ -2x+6 = x ]
Now, substitute the value of x in either of the 2 equations and you'll get the y coordinate for the point.

Does that help ?
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Re: Point P(a,b) is randomly selected in the region enclosed by  [#permalink]

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New post 23 Feb 2017, 22:59
Shruti0805 wrote:
GovindAgrawal wrote:
how do we get this point of intersection?


Hi,

Let me try explaining. It took me a while figuring it out myself.

First, we can replace the lines x=0 and y=0 by a line x=y. That leaves us with just 2 lines :- 1st is x=y and 2nd is 3y= -2x + 6.

To find the intersection of these 2 lines, equate the RHS variables and you'll get x=6/5. [ -2x+6 = x ]
Now, substitute the value of x in either of the 2 equations and you'll get the y coordinate for the point.

Does that help ?


Just a little clarification.

We do NOT have to replace lines x=0 and y=0 with x=y

Instead we need to understand that y and x co ordinate are equL on line y=x I.e. on that line a=B

But since we need b>a so we need all points above line y=x which are bounded by other three lines which are
x=0
y=0
And -2x+6 = 3x

I hope this helps
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Re: Point P(a,b) is randomly selected in the region enclosed by  [#permalink]

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New post 02 Aug 2018, 05:44
Bunuel , why we can not find the area of the triangle in yellow through 90-45-45 form?
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Re: Point P(a,b) is randomly selected in the region enclosed by &nbs [#permalink] 02 Aug 2018, 05:44
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