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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) \(\frac{5}{72}\)

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08 Feb 2013, 12:03

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mikemcgarry wrote:

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) \(\frac{5}{72}\)

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

Mike

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!
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09 Feb 2013, 18:25

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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

Re: Region R is a square in the x-y plane with vertices [#permalink]

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05 Mar 2013, 11:26

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.

Re: Region R is a square in the x-y plane with vertices [#permalink]

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03 May 2013, 23:32

mikemcgarry wrote:

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) \(\frac{5}{72}\)

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

Mike

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?

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) \(\frac{5}{72}\)

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

Mike

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

Hi Tarun,

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

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22 Aug 2014, 15:42

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08 Dec 2015, 10:44

Hello from the GMAT Club BumpBot!

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