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655-705 Level|   Geometry|   Probability|      
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apoorvasrivastva
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 16 Nov 2009, 07:27
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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 x-y>0 ?

A. 1/5
B. 1/3
C. 1/2
D. 2/3
E. 4/5
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E
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Bunuel
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 16 Nov 2009, 07:50
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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 x-y>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.

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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 24 Dec 2009, 08:35
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Check other Probability and Geometry questions in our Special Questions Directory.

GEOMETRY: Shaded Region Problems!

22. Probability



For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread


Hope it helps.
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 15 Mar 2013, 14:55
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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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apoorvasrivastva
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 16 Nov 2009, 09:31
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Bunuel


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 are=4*4/2=8.

P=8/10=4/5

Answer: E.
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If i go by the way u say....
shouldn't the probability be <4/5?

probability = area of region where x-y>0 / total are of region = x/0.5*4*5 = x/10

point (4,4) is where x-y=0. point(4, <4) is where x-y>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 x-y is not >0. any value of y below 4 would give x-y>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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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 16 Nov 2009, 12:26
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Also if i go by the way u say....
shouldn't the probability be <4/5?

probability = area of region where x-y>0 / total are of region = x/0.5*4*5 = x/10

point (4,4) is where x-y=0. point(4, <4) is where x-y>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 x-y is not >0. any value of y below 4 would give x-y>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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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 x-y > 0 or x-y > 0. Note that the region where x-y=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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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 03 Jan 2013, 01:31
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sagarsabnis
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 x-y>0 ?

A. 1/5
B. 1/3
c. 1/2
D. 2/3
E. 4/5

please help with this one.
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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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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 26 Mar 2013, 08:55
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johnwesley
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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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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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 27 Mar 2013, 08:47
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perilm41
johnwesley
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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A point by definition has no length, area or any other dimension, thus this won't affect the answer.

Check this discussion for more: a-5-meter-long-wire-is-cut-into-two-pieces-if-the-longer-106448.html
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 25 Jul 2013, 00:10
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Bunuel

Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4).
Answer: E.
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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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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 02 Sep 2013, 14:15
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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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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 https://gmatclub.com/blog/2013/02/geomet ... -the-gmat/

apoorvasrivastva
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 x-y>0 ?

A. 1/5
B. 1/3
c. 1/2
D. 2/3
E. 4/5
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 19 Aug 2017, 10:54
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Bunuel
sagarsabnis
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 x-y>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 --> 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) (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.
Show more

Just out of curiosity!

In the above case, what would be the probability that x-y<0 and that x=y?

I think P(x-y<0)=0.2, but don't know exactly the P(x=y). It may be 0, but don't know why.

P(x-y>0)=0.8
P(x-y<0)=0.2
P(x=y)=0
Total 1.0

Would you help me?
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 19 Aug 2017, 11:03
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Bunuel
sagarsabnis
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 x-y>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 --> 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) (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 x-y<0 and that x=y?

I think P(x-y<0)=0.2, but don't know exactly the P(x=y). It may be 0, but don't know why.

P(x-y>0)=0.8
P(x-y<0)=0.2
P(x=y)=0
Total 1.0

Would you help me?
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A line segment has only one dimension, length, no area. So, P(x=y) = 0.
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 23 Feb 2021, 09:51
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Bunuel
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 x-y>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.

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Of course after Bunuel answers it becomes exceedingly clear. However, perhaps this is too rudimentary...what triggered you to use y = x? That's the key to this question and I didn't recognize that it's needed to answer. Is there a 'general rule' for when we should use it?
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 01 Dec 2022, 09:04
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My logic is different but answer is correct.

I used 1-P(not) so taking all points that don't satisfy x-y>0

0,0 1,1 2,2 3,3 4,4 and 4,5
= 6 points
And all points on the plane

6x5 = 30 points

Applied 1- 6/30

Answer = 4/5

If this method is right bb let me know

Posted from my mobile device
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 12 May 2023, 04:46
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Bunuel
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 x-y>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.

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Untitled.png
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I have one doubt , it is given that x> y and ,as per the explanation, favorable region has vertexes at (0,0) (0,4) and (4,4). So, the randomly picked point can be (4,4) then doesn't this violate the condition x>y?
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 12 May 2023, 06:14
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Bunuel
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 x-y>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.

Show Spoiler
Attachment:
Untitled.png


I have one doubt , it is given that x> y and ,as per the explanation, favorable region has vertexes at (0,0) (0,4) and (4,4). So, the randomly picked point can be (4,4) then doesn't this violate the condition x>y?
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Your doubt is already addressed here: https://gmatclub.com/forum/in-the-xy-pl ... l#p1203738
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a 06 Jul 2025, 12:43
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