Region R is a square in the x-y plane with vertices

Question

Region R is a square in the x-y plane with vertices J = (–1, –2), K = (–1, 4), L = (5, 4), and M = (5, –2). What is the probability that a randomly selected point in region R lies below the line 3x – 5y = 10?
(A) \(\frac{5}{12}\)
(B) \(\frac{5}{18}\)
(C) \(\frac{5}{24}\)
(D) \(\frac{5}{36}\)
(E) \(\frac{5}{72}\)

For thoughts on geometric probability, as well as a complete solution with diagram to this particular problem, see:
https://magoosh.com/gmat/2013/geometric- ... -the-gmat/

Experts: are there any words of wisdom you would like to share on the topic of Geometric Probability?

Mike :-)

nave · Accepted Answer

The square is formed by x=5 & x=-1 ; y=4 & y=-2

-1 to 5 or -2 to 4 = 6 which is the length of the square (if it was an rectangle we would have to find the length and breadth)

So the area of the square is 36.

To find the points where the line 3x-5y=10 cuts across the square, insert the two values of x or y into the equation

x=5 
=> 3*5-5y=10 
=> y=1. So (5,1) is first point of intersection of the line with the square.

x=-1
=> 3*(-1)-5y=10
=> y=-(7/5). This point (-1, -7/5) is not on any of the boundary of the square so this is not relevant. Let's try inserting the values of y to find where is the second point of intersection is

y=-2
=> 3x-5*(-2)=10
=> x=0. So (0, -2) is the second point of intersection of the line with the square.

So the line cuts the square at two adjacent sides of the square to form a triangle with base 5 and height 3. Area of this triangle is 15/2

Probability of point lying in this triangle = (15/2)/36 = 5/24

Vips0000 · Answer

Basically the square is formed with 4 lines x=-1, x=5, y=-2,y=4
Area of square =36
Now line is given as 5y=3x-10
We can see that the points on which this line intersects 4 lines forming square are (0,-2) and (5,1)
Thus area of triangle formed under the line =15/2
Probability of a point in this area=(15/2)/36 =5/24
Ans C it is!

shreerajp99 · Answer

1 doubt i have here is that if we try to find x intercept of that line,we get x =10/3 but as per the discussion here,the line intercepts x axis at 5,1.

Thanks,
Shreeraj

kuttingchai · Answer

will vote for C

Area of square = 36

line passes thr' = (0,-2) & (0.3.3)

Line intersect the square @ = (0,-2) & (5,1) - this forms a triangle with base=5 and height = 3

Area of triangle = 15/2

probability = 15/2 * 1/36 = 5 / 24

tarunjagtap · Answer

Hi,

i might be wrong in here, please correct me.

prob = area of triangle / total square area.
but in the question it is What is the probability that a randomly selected point in region R lies   below   the line 3x – 5y = 10?

so shouldn't the answer be 1 - (5/24)

thanks in advance

Narenn · Answer

Hi Tarun,

Go thru the attached figure. then you will come to know what are you missing exactly.

Narenn

tarunjagtap · Answer

oh ok .. got it.. thanks Narenn.. I dont know what was going through my mind..

actually i plotted the line wrongly.. hence the stupid question.

Ashishsteag · Answer

X=5 is the equation of the line LM
3x-5y=10 intersects y axis at -2 and x=5 at (5,1).

So,we get a triangle whose area=1/2*5*3=15/2
Now Area of square=6^2=36

So,probability=15/2/36=15/72=5/24

avaniguptaonly · Answer

How do we get x co-ordinate as 5 ?

Experts pls explain  Bunuel   GMATNinja

Bunuel · Answer

The x-coordinate of an intersection point of any line that crosses the line  x = 5  will have the x-coordinate of 5. After all, if it's on the line x = 5, how could its x-coordinate be anything else? Or were you asking something else?

avaniguptaonly · Answer

Oh ok! Got it. Thanks. My bad

Nabneet · Answer

Bunuel .

I have the same doubt.

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Region R is a square in the x-y plane with vertices

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Region R is a square in the x-y plane with vertices

Prep Toolkit

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