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Point P(a,b) is randomly selected in the region enclosed by
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Updated on: 27 Sep 2013, 00:35
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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?
Originally posted by jellybean23 on 26 Sep 2013, 22:51.
Last edited by Bunuel on 27 Sep 2013, 00:35, 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
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27 Sep 2013, 00:57
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 [ 13.38 KiB  Viewed 14358 times ]
The area of the triangle enclosed by 3y + 2x = 6 (y=22/3*x), x=0 and y = 0 is 1/2*2*3=3. We need the ycoordinate of P to be greater than its xcoordinate (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: settconsistsofallpointsxysuchthatx2y21if15626.htmlinthecoordinateplanerectangularregionrhasverticesa104869.htmlinthexyplaneatrianglehasvertexes0040and86926.htmlinthexyplaneatrianglehasvertexes0040and88395.htmlHope it helps.
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Re: Point P(a,b) is randomly selected in the region enclosed by
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20 Jul 2014, 09:12
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
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20 Jul 2014, 10:33
ronr34 wrote: Hi Bunuel, Can you explain how you calculated the area of the triangle? Look at the diagram below: Attachment:
Untitled.png [ 13.67 KiB  Viewed 13661 times ]
The base of the triangle is 2 (vertical green line) and the height of the triangle is 6/5 (horizontal green line, xcoordinate 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
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31 Jul 2016, 03:10
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=22/3*x), x=0 and y = 0 is 1/2*2*3=3. We need the ycoordinate of P to be greater than its xcoordinate (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: settconsistsofallpointsxysuchthatx2y21if15626.htmlinthecoordinateplanerectangularregionrhasverticesa104869.htmlinthexyplaneatrianglehasvertexes0040and86926.htmlinthexyplaneatrianglehasvertexes0040and88395.htmlHope 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
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27 Jan 2017, 01:27
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
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23 Feb 2017, 21:52
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
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23 Feb 2017, 21:59
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
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02 Aug 2018, 04:44
Bunuel , why we can not find the area of the triangle in yellow through 904545 form?
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