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Manager  Joined: 29 Jul 2009
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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Question Stats: 58% (01:59) correct 42% (02:29) wrong based on 431 sessions

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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
Math Expert V
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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Manager  Joined: 22 Jul 2009
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In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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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
##### General Discussion
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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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.

$$P=\frac{8}{10}=\frac{4}{5}$$.

Attachment: Untitled.png [ 13.33 KiB | Viewed 11641 times ]

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Manager  Joined: 29 Jul 2009
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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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 --> 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

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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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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apoorvasrivastva wrote:

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...

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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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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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 x-y>0 ?

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

OA is 4/5

x - y > 0 or x > y is possible below the line x = y in the given triangle / co-ordinate 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 xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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GMAT TIGER wrote:
x - y > 0 or x > y is possible below the line x = y in the given triangle / co-ordinate 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,  [#permalink]

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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 x-y>0 ?

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

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

Detailed Solution: Line x=y as a Boundary
Manager  Joined: 24 Jan 2013
Posts: 64
Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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16
3
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
Attachments

File comment: Drawing resolution Drawing resolution.jpg [ 13.41 KiB | Viewed 13376 times ]

Intern  Joined: 17 Nov 2012
Posts: 1
Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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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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Joined: 02 Sep 2009
Posts: 59727
Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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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: a-5-meter-long-wire-is-cut-into-two-pieces-if-the-longer-106448.html
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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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Bunuel wrote:
Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4).

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 xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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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.

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 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  [#permalink]

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Taking the third vertex as (4,4) means, to me, x=y not x>y. am I wrong?
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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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  [#permalink]

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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 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) (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

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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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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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 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) (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

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?

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  [#permalink]

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Ans is shown in picture as part of my calculation
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File comment: Clearly ans is E IMG_4495.JPG [ 1.42 MiB | Viewed 6820 times ]

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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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+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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