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What is the probability the x-y>0 ?

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apoorva.srivastva Master | Next Rank: 500 Posts Default Avatar
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What is the probability the x-y>0 ?

Post Mon Nov 16, 2009 6:19 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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 Default Avatar
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    Post 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?

    apoorva.srivastva Master | Next Rank: 500 Posts Default Avatar
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    Post 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

    apoorva.srivastva Master | Next Rank: 500 Posts Default Avatar
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    Post 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

    mridul_dave Senior | Next Rank: 100 Posts Default Avatar
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    Post 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.

    Thanked by: apoorva.srivastva
    apoorva.srivastva Master | Next Rank: 500 Posts Default Avatar
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    Post 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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    Post 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.

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    mridul_dave Senior | Next Rank: 100 Posts Default Avatar
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    Post 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 ?

    Post Wed Apr 17, 2013 9:34 am
    Quote:
    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 Default Avatar
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    Post 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 ½ which is 2/3

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