What is the probability the x-y>0 ?

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apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

by apoorva.srivastva » Mon Nov 16, 2009 6:19 am

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..

[spoiler ] OA is 4/5 [/spoiler]

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life is a test: Master | Next Rank: 500 Posts; Posts: 138; Joined: Mon Mar 02, 2009 12:02 pm; Thanked: 15 times

by life is a test » Mon Nov 16, 2009 8:23 am

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...

anyone else have any thoughts on this?

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apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

by apoorva.srivastva » Mon Nov 16, 2009 8:39 am

life is a test wrote: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...

anyone else have any thoughts on this?

precisely....what i meant to say was the probabilty has to be less than 4/5...for the reason pointed abpove in red

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

Last edited by apoorva.srivastva on Mon Nov 16, 2009 9:08 am, edited 1 time in total.

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apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

by apoorva.srivastva » Mon Nov 16, 2009 8:50 am

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

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mridul_dave: Senior | Next Rank: 100 Posts; Posts: 41; Joined: Tue Oct 06, 2009 9:49 am; Thanked: 4 times

by mridul_dave » Mon Nov 16, 2009 10:22 am

apoorva.srivastva 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..

[spoiler ] OA is 4/5 [/spoiler]

Here is how I did it.
make a quick sketch of the triangle and make a line that represents x = y. This line joins origin (0.0) and a point (4,4) on the triangle, making another triangle inside the orginal one.
All points in the inside triangle satisfy x-y > 0. ( Think x > y )

Probability that we need is area of inner triangle / area of outer triangle.

you will get 4/5 exact. Hence the answer.

This covers integers and non integer values of x and y.
Please let me know if you have any questions.

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apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

by apoorva.srivastva » Mon Nov 16, 2009 10:45 am

[quote="mridul_dave]
Here is how I did it.
make a quick sketch of the triangle and make a line that represents x = y. This line joins origin (0.0) and a point (4,4) on the triangle, making another triangle inside the orginal one.
All points in the inside triangle satisfy x-y > 0. ( Think x > y )

Probability that we need is area of inner triangle / area of outer triangle.

you will get 4/5 exact. Hence the answer.

This covers integers and non integer values of x and y.
Please let me know if you have any questions.[/quote]

yeh mate thanks ..cool explanation!!

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by Ian Stewart » Mon Nov 16, 2009 11:20 am

life is a test wrote: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...

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.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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mridul_dave: Senior | Next Rank: 100 Posts; Posts: 41; Joined: Tue Oct 06, 2009 9:49 am; Thanked: 4 times

by mridul_dave » Thu Nov 19, 2009 6:55 pm

Ian Stewart wrote:
life is a test wrote: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...
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.

Ian,

With the logic of area of the line being zero, it cannot mean that probability of x = y is 0. I had decided not to think about it as its just GMAT. But still... what is the probability that x = y ?

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Apr 17, 2013 9:34 am

In the xy-plane, a triangle has vertices (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 received a PM requesting that I post an explanation.

Question rephrased: In what portion of the triangle is y<x?

Area of the whole triangle = (1/2)*4*5 = 10.
The portion below y=x is where y<x.
This region is the triangle with vertices at (0,0), (4,0) and (4,4).
Area of this triangle = (1/2)*4*4 = 8.
(Portion below y=x)/(Total area) = 8/10 = 4/5.

The correct answer is E.

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vipulgoyal: Master | Next Rank: 500 Posts; Posts: 468; Joined: Mon Jul 25, 2011 10:20 pmFollowed by: 4 members; Thanked: 29 times

by vipulgoyal » Wed Apr 17, 2013 10:31 pm

Experts please suggest

I don't understand why in the required probability everyone considering triangle with vertices (0,0), (4,0) and (4,4).rather considering trapezoid with vertices at (0,0), (4,0) and (4,4). And one on diagonal unknown, because there are points which are in between diagonals of two triangle below point( 4,4) which are not covered in the probability

In my opinion probability should be less then 4/5 and greater then Â½ which is 2/3

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