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# In the xy-plane, a triangle has vertexes (0,0), (4,0) and

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

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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
[Reveal] Spoiler: OA

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

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16 Nov 2009, 07:50
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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

I am getting 3/5 as the answer..please explain the reasoning..

[Reveal] Spoiler:
OA is 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.

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

[Reveal] Spoiler:
Attachment:

Untitled.png [ 13.33 KiB | Viewed 4549 times ]

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

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16 Nov 2009, 09:24
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

well my reasoning goes like this..... P(e) = F(e) / T(e)

T(e) = 4C1 ( ways of selecting any x co-ordinate in the region ) * 5C1 ( ways of selecting any y co ordinate in the region)
= 4*5= 20

for F(e) we are required to find x>y .... so for this X co-ordinate can be chosen in 4 ways and the y co-ordinate can be chosen in 3 ways to satisfy the x>y condition

so T(e) 4*3=12

so p(e) = 12/20 = 3/5

i am considering only integers is this the problem???
kindly guide

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

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16 Nov 2009, 09:28
apoorvasrivastva wrote:
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

well my reasoning goes like this..... P(e) = F(e) / T(e)

T(e) = 4C1 ( ways of selecting any x co-ordinate in the region ) * 5C1 ( ways of selecting any y co ordinate in the region)
= 4*5= 20

for F(e) we are required to find x>y .... so for this X co-ordinate can be chosen in 4 ways and the y co-ordinate can be chosen in 3 ways to satisfy the x>y condition

so T(e) 4*3=12

so p(e) = 12/20 = 3/5

i am considering only integers is this the problem???
kindly guide

T(e) = 4C1 ( ways of selecting any x co-ordinate in the region ) * 5C1 ( ways of selecting any y co ordinate in the region)
= 4*5= 20

this would make the region a rectangle (4 x 5) but you need to multiply by 1/2 since it's a right triangle

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

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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 --> 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

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

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16 Nov 2009, 09:32
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for F(e) we are required to find x>y .... so for this X co-ordinate can be chosen in 4 ways and the y co-ordinate can be chosen in 3 ways to satisfy the x>y condition

the y can be chosen in 4 ways as well (4 x 4) * 1/2 (for a triangle) = 8

think of the y = x coordinate

0,0
1,1
2,2
3,3
4,4

anything below this line satisfies the equation 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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16 Nov 2009, 09:53
lagomez wrote:
for F(e) we are required to find x>y .... so for this X co-ordinate can be chosen in 4 ways and the y co-ordinate can be chosen in 3 ways to satisfy the x>y condition

the y can be chosen in 4 ways as well (4 x 4) * 1/2 (for a triangle) = 8

think of the y = x coordinate

0,0
1,1
2,2
3,3
4,4

anything below this line satisfies the equation x>y

ok now i get it!!!!

The probability for x>y is nothing but = 1 - P(y>x) = 1- 1/5 = 4/5

reasoning as follows

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

x = 0.5 * 4* 1 = 2

so P (y>x) = 2/10= 1/5

so P(x>y) = 1- (1/5)
= 4/5

PLEASE CORRECT ME IF I AM WRONG!!!

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

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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 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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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 x-y>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..

[Reveal] Spoiler:
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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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 / 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, a triangle has vertexes (0,0), (4,0) and [#permalink]

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17 Nov 2009, 14:30
Bunuel wrote:
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.

Thanks Bunuel. You are amezing..

Updated ...taking advantage of being a moderator.
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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]

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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).

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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02 Sep 2013, 13:21
Here is how I thought of the problem.
Considering only integers, the X values can be 1,2,3 or 4.
For either of these four values, in order to get X-Y > 0, Y values have to be one of (0,1,2 or 3)
Probability of both of these events is (4/5)*(4/4). Hence I got the value 4/5

idk if this worked out only because of a fluke or is this wrong?

Thanks

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

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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.

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

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20 Jul 2014, 06:28
Bunuel wrote:
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

I am getting 3/5 as the answer..please explain the reasoning..

[Reveal] Spoiler:
OA is 4/5

Attachment:
Untitled.png

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

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

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

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

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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?

Bunuel wrote:
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

I am getting 3/5 as the answer..please explain the reasoning..

[Reveal] Spoiler:
OA is 4/5

Attachment:
Untitled.png

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

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

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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?

Bunuel wrote:
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

I am getting 3/5 as the answer..please explain the reasoning..

[Reveal] Spoiler:
OA is 4/5

Attachment:
Untitled.png

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

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] 20 Jul 2017, 20:51
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