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]
What is the probability the x-y>0 ?
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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?
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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precisely....what i meant to say was the probabilty has to be less than 4/5...for the reason pointed abpove in redlife 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?
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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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
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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Here is how I did it.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]
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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[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!!
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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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.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...
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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Ian Stewart wrote: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.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...
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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I received a PM requesting that I post an explanation.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
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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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 ½ which is 2/3
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 ½ which is 2/3