In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a

Question

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

Bunuel · Accepted Answer

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.

\(P=\frac{8}{10}=\frac{4}{5}\).

Answer: E.

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Bunuel · Answer

Check other  Probability and Geometry  questions in our  Special Questions Directory .

GEOMETRY: Shaded Region Problems!

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For more:
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Hope it helps.

johnwesley · Answer

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

apoorvasrivastva · Answer

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...

IanStewart · Answer

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.

mbaiseasy · Answer

My approach is to use the x=y line which is the boundary that divides x>y and 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 Answer: E Detailed Solution: Line x=y as a Boundary

perilm41 · Answer

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)?

Bunuel · Answer

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

Jaisri · Answer

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).

Narenn · Answer

Yes, The points fall below the line y = x will give us x > y and hence it is the favorable region. Refer this article for more info on Geometric Probabilities https://gmatclub.com/blog/2013/02/geomet ... -the-gmat/ https://lh5.googleusercontent.com/-_krGIu8H_TU/UiUAbqHjpXI/AAAAAAAAAJ0/DvVj1SWvW2Q/w967-h566-no/Drawing1-Model.png

Mahmud6 · Answer

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?

Bunuel · Answer

A line segment has only one dimension, length, no area. So, P(x=y) = 0.

CEdward · Answer

Of course after Bunuel answers it becomes exceedingly clear. However, perhaps this is too rudimentary...what triggered you to use y = x? That's the key to this question and I didn't recognize that it's needed to answer. Is there a 'general rule' for when we should use it?

ShreyashGupta188 · Answer

My logic is different but answer is correct.

I used 1-P(not) so taking all points that don't satisfy x-y>0

0,0 1,1 2,2 3,3 4,4 and 4,5
= 6 points 
And all points on the plane

6x5 = 30 points

Applied 1- 6/30

Answer = 4/5

If this method is right  bb  let me know

Posted from my mobile device

akt715 · Answer

I have one doubt , it is given that x> y and ,as per the explanation, favorable region has vertexes at (0,0) (0,4) and (4,4). So, the randomly picked point can be (4,4) then doesn't this violate the condition x>y?

Bunuel · Answer

Your doubt is already addressed here: https://gmatclub.com/forum/in-the-xy-pl ... l#p1203738

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