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In the xyplane, a triangle has vertexes (0,0), (4,0) and
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16 Nov 2009, 07:27
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In the xyplane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that xy>0 ? A. 1/5 B. 1/3 C. 1/2 D. 2/3 E. 4/5
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In the xyplane, a triangle has vertexes (0,0), (4,0) and
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24 Dec 2009, 08:35




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In the xy plane, a triangle has vertexes (0,0), (4,0) and
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24 Dec 2009, 08:09
In the xy plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that xy>0 ?
A. 1/5 B. 1/3 C. 1/2 D. 2/3 E. 4/5




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In the xyplane, a triangle has vertexes (0,0), (4,0) and
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16 Nov 2009, 07:50
In the xy plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that xy>0 ? A. 1/5 B. 1/3 c. 1/2 D. 2/3 E. 4/5 We have right triangle with the area = 4*5/2 = 10. Consider the line y<x. All the points which satisfy this equation (are below the line y = x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x > y > x  y > 0). The probability that the point will be from this region is: (Area of this region)/(Area of the triangle). Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). As y = x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable area = 4*4/2 = 8. \(P=\frac{8}{10}=\frac{4}{5}\). Answer: E. Attachment:
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In the xyplane, a triangle has vertexes (0,0), (4,0) and
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16 Nov 2009, 09:31
Bunuel wrote: We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y > xy>0).
The probability that the point will be from this region is: Area of this region/Area of the triangle.
Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8.
P=8/10=4/5
Answer: E.
If i go by the way u say.... shouldn't the probability be <4/5? probability = area of region where xy>0 / total are of region = x/0.5*4*5 = x/10 point (4,4) is where xy=0. point(4, <4) is where xy>0. x = 0.5 * 4*(<4) = <8 hence x/10 = <8/10 = <4/5 the point here is that to get 4/5 you would have to take the region where x and y = 4 in which case xy is not >0. any value of y below 4 would give xy>0 hence the probability should be less than 4/5...I dont think you can place an exact value on the probability as x and y can be decimals...



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Re: In the xyplane, a triangle has vertexes (0,0), (4,0) and
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16 Nov 2009, 12:26
apoorvasrivastva wrote: Also if i go by the way u say.... shouldn't the probability be <4/5?
probability = area of region where xy>0 / total are of region = x/0.5*4*5 = x/10
point (4,4) is where xy=0. point(4, <4) is where xy>0.
x = 0.5 * 4*(<4) = <8 hence x/10 = <8/10 = <4/5
the point here is that to get 4/5 you would have to take the region where x and y = 4 in which case xy is not >0. any value of y below 4 would give xy>0 hence the probability should be less than 4/5...I dont think you can place an exact value on the probability as x and y can be decimals... I responded to this same post on BTG, but I'll paste that here: To respond specifically to the concern above: the answer is still exactly 4/5, regardless of whether the question says xy > 0 or xy > 0. Note that the region where xy=0 is just the line x=y; its area is zero. So you can subtract its area in your calculation if you like, but it won't change the answer.
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Re: In the xyplane, a triangle has vertexes (0,0), (4,0) and
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16 Nov 2009, 15:39
apoorvasrivastva wrote: In the xy plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probabilty that xy>0 ? A. 1/5 B. 1/3 c. 1/2 D. 2/3 E. 4/5 I am getting 3/5 as the answer..please explain the reasoning.. x  y > 0 or x > y is possible below the line x = y in the given triangle / coordinate system. i.e. the area below x = y line in the triangle is the prob that x>y. So draw a line from origin to the point (4, 4). This makes a triangle with vertexes (0,0), (4,0) and (4,4). Given triangle with vertexes (0,0), (4,0) and (4,5) has an area of 10. The area of the triangle with vertexes (0,0), (4,0) and (4,4) is 8. Area of new triangle/Area of the original triangle = 8/10 = 4/5. This is the prob that x>y.
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Re: In the xyplane, a triangle has vertexes (0,0), (4,0) and
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16 Nov 2009, 15:48
GMAT TIGER wrote: x  y > 0 or x > y is possible below the line x = y in the given triangle / coordinate system. i.e. the area below x = y line in the given triangle is the prob that x>y. So draw a line from origin to the point (4, 4). This line makes a triangle with vertexes (0,0), (4,0) and (4,4).
Given triangle with vertexes (0,0), (4,0) and (4,5) has an area of 20. The area of the triangle with vertexes (0,0), (4,0) and (4,4) is 16.
Area of new triangle/Area of the original triangle = 16/20 = 4/5. This is the prob that x>y. Areas are 10 and 8 accordingly. You forgot to divide by 2 in both cases. Though it didn't affect the final answer.
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Re: In the xy plane,
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03 Jan 2013, 01:31
sagarsabnis wrote: In the xy plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that xy>0 ?
A. 1/5 B. 1/3 c. 1/2 D. 2/3 E. 4/5
please help with this one. My approach is to use the x=y line which is the boundary that divides x>y and x<y coordinate system. Below the x=y line is the region of points with x > y. Above the x=y line is the region of points with x < y.Get the area of the given triangle: (4)(5) / 2 = 10 Get the area of the smaller triangle drawn below the line x=y which has (0,0), (4,0) and (4,4) for its coordinates: (4)(4)/2 = 8 P = 8/10 = 4/5 Answer: E Detailed Solution: Line x=y as a Boundary
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Re: In the xy plane, a triangle has vertexes (0,0), (4,0) and
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15 Mar 2013, 14:55
See attached drawing. The probability will be the area of the dark triangle divided by the area of the original triangle: Area dark triangle (where X>Y is true): \(\frac{1}{2}*4*4\) Area original triangle: \(\frac{1}{2}*4*5\) \(Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}\) Solution E
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Re: In the xy plane, a triangle has vertexes (0,0), (4,0) and
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26 Mar 2013, 08:55
johnwesley wrote: See attached drawing.
The probability will be the area of the dark triangle divided by the area of the original triangle:
Area dark triangle (where X>Y is true): \(\frac{1}{2}*4*4\)
Area original triangle: \(\frac{1}{2}*4*5\)
\(Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}\)
Solution E What about (4,4)? that is in the shaded region, but is not greater than 0 (or algebraically x=4 is not greater than y=4)?



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Re: In the xy plane, a triangle has vertexes (0,0), (4,0) and
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27 Mar 2013, 08:47
perilm41 wrote: johnwesley wrote: See attached drawing.
The probability will be the area of the dark triangle divided by the area of the original triangle:
Area dark triangle (where X>Y is true): \(\frac{1}{2}*4*4\)
Area original triangle: \(\frac{1}{2}*4*5\)
\(Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}\)
Solution E What about (4,4)? that is in the shaded region, but is not greater than 0 (or algebraically x=4 is not greater than y=4)? A point by definition has no length, area or any other dimension, thus this won't affect the answer. Check this discussion for more: a5meterlongwireiscutintotwopiecesifthelonger106448.html
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Re: In the xyplane, a triangle has vertexes (0,0), (4,0) and
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25 Jul 2013, 00:10
Bunuel wrote: Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). Answer: E. Hi Bunuel, Since we have to select a point within the given triangular region formed by (0,0(, (4,0) and (4,5), isn't the favorable region formed by (0,0), (4,4) and (4,5)? If not, kindly clarify how the vertexes for favorable region would be (0,0) (0,4) and (4,4).



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Re: In the xyplane, a triangle has vertexes (0,0), (4,0) and
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02 Sep 2013, 14:15
Jaisri wrote: Hi Bunuel, Since we have to select a point within the given triangular region formed by (0,0(, (4,0) and (4,5), isn't the favorable region formed by (0,0), (4,4) and (4,5)? If not, kindly clarify how the vertexes for favorable region would be (0,0) (0,4) and (4,4). Yes, The points fall below the line y = x will give us x > y and hence it is the favorable region. Refer this article for more info on Geometric Probabilities http://gmatclub.com/blog/2013/02/geomet ... thegmat/apoorvasrivastva wrote: In the xyplane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that xy>0 ?
A. 1/5 B. 1/3 c. 1/2 D. 2/3 E. 4/5
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In the xyplane, a triangle has vertexes (0,0), (4,0) and
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20 Jul 2017, 17:08
Taking the third vertex as (4,4) means, to me, x=y not x>y. am I wrong?



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In the xyplane, a triangle has vertexes (0,0), (4,0) and
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20 Jul 2017, 20:51
Yoohanna wrote: Taking the third vertex as (4,4) means, to me, x=y not x>y. am I wrong? A point has no dimension, so including (4, 4) won't affect the probability.
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Re: In the xy plane, a triangle has vertexes (0,0), (4,0) and
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19 Aug 2017, 10:54
Bunuel wrote: sagarsabnis wrote: In the xy plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that xy>0 ?
A. 1/5 B. 1/3 c. 1/2 D. 2/3 E. 4/5
please help with this one. We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y > xy>0). The probability that the point will be from this region is: Area of this region/Area of the triangle. Favorable region is also right triangle with vertexes at (0,0) (4,0) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8. P=8/10=4/5 Answer: E. Just out of curiosity! In the above case, what would be the probability that xy<0 and that x=y? I think P(xy<0)=0.2, but don't know exactly the P(x=y). It may be 0, but don't know why. P(xy>0)=0.8 P(xy<0)=0.2 P(x=y)=0 Total 1.0 Would you help me?
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Re: In the xy plane, a triangle has vertexes (0,0), (4,0) and
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19 Aug 2017, 11:03
Mahmud6 wrote: Bunuel wrote: sagarsabnis wrote: In the xy plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that xy>0 ?
A. 1/5 B. 1/3 c. 1/2 D. 2/3 E. 4/5
please help with this one. We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y > xy>0). The probability that the point will be from this region is: Area of this region/Area of the triangle. Favorable region is also right triangle with vertexes at (0,0) (4,0) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8. P=8/10=4/5 Answer: E. Just out of curiosity! In the above case, what would be the probability that xy<0 and that x=y? I think P(xy<0)=0.2, but don't know exactly the P(x=y). It may be 0, but don't know why. P(xy>0)=0.8 P(xy<0)=0.2 P(x=y)=0 Total 1.0 Would you help me? A line segment has only one dimension, length, no area. So, P(x=y) = 0.
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Re: In the xy plane, a triangle has vertexes (0,0), (4,0) and
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19 Aug 2017, 11:13
Ans is shown in picture as part of my calculation Answer is E
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Re: In the xyplane, a triangle has vertexes (0,0), (4,0) and
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18 Apr 2018, 09:11
+1 for option E. A great question combining the concepts of probability and coordinate geometry. Draw the graph and the required probability will be the ratio of two areas. 8/10 i.e, 4/5 or option E.
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