The thought was this:



Any line passing through the point P which is at 6 on the x axis will pass through one of the 4 equal dotted regions (except x axis). Out of these, lines passing through 2 of the regions are not acceptable (red lines) while those passing through other 2 are acceptable (green lines).
Hence, 1/2 of the region is fine.

Probability that the line does not pass through ABCD is 1/2.
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I am responding to a pm from himanshuhpr.

First of all, I am intrigued by the solution of the very intelligent VeritasPrepKarishma, for whom I have considerable respect. Nevertheless, I beg to differ.

I believe there are some significant flaws in the question. First of all, the grammar is embarrassingly atrocious, but we'll let that pass. More to the point === we are told that the program "randomly" generates lines --- how exactly is this process randomized? For example, the line goes through P --- Are all slopes equally likely? Or are all angles equally likely? If we select P and a y-intercept, are all y-intercept equally likely? Are we randomly selecting one other point in the x-y plane, and all of those points are equally likely? Those are four different ways of specifying the probabilistic process, and I believe at least some of them give different numerical answer to the question. As soon as the thing you are specifying (here, a line) can be specified in a variety of ways, its not enough to say "random" --- you have to specify exactly how the selection process proceeds.

Because the question does not specify the details of the probabilistic selection process, I believe it is a fundamentally flawed question.

Does that make sense?

Mike
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Mike , Thanks for replying.

I somewhat understand the flaw your are trying to point out but given the question in this ' FLAWED ' form what should be the best possible response/ method to solve the above.

Dear himanshuhpr,

I understand your desire to understand. As I explained in my pm response, I think this question is too flawed to admitted of an unambiguously good solution.

Instead, I created a similar question, and I believe in this question the entirely selection process of the line is very tightly defined, making the question unambiguous. Here's where it's posted ......

hard-probability-problem-lines-in-the-x-y-plane-142846.html

After some folks wrestle with it, I will post a solution.

I hope that's helpful.

Mike
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I think the question seems to be fine and solution that Karishma provided is excellent.

If one can notice in original picture posted, cooridnates of point C ,D and P are such that angle DPC is 90. (This can be verified using coordinates or slopes)
Thus, for any line of the form y=mx+b, passing through point P. There is a 90' region from which it can not pass, else it will go inside the square ABCD. Outside of this 90 it can pass wherever it wants. it will be ok.

There is overall 180' region, starting from x-asis and going in anticlock wise, where this line can be drawn. (considering only above x-axis - as once you rotate line enough to go below x-axis, the other end of line would be above x-axis)

Thus probability = Total favourable area/Total area = 1/2

Now, coming to how it is 90' and 180'. We can consider this solution only in first quadrant or in first & fourth quadrant both. However when we consider both quadrants, since it will provide a mirror image across x-axis.
total area, which is not desired, formed by line = 90*2 and total area possible for line = 180*2 = 360

still you get same ans 1/2.

Interesting... while I do agree the question is poorly worded. I was able to come to (C) as the answer, but I had to recall my SIN/COS/TAN angles...which seems overkill for the GMAT. Even for a 700+ question...

Step 1:
The graph shows point P(6 units, 0 units), using either points C(4 units, 2 units) or D (8 units, 2 units) you can derive a triangle with two sides of length 2 units, and a hypotenuse we don't need to know.

Remembering SOH/CAH/TOA -> Use TOA. Tan of Angle (Degrees) = Opposite Length / Adj. Length = 2 / 2 = 1. Tan of 45 Degrees = 1.

Therefore, the sector bounded by the two linear equations is 180 - (45 x 2) = 90 degrees.

Step 2:
Realize that it will not simply be 90 / 360 (= 1 / 4) as this will include linear equations that cross ABCD.

Using the two bounded lines, visualize a North/South/East/West quadrant and realize that North/South sectors are covered by the same line equations, East/West sectors are covered by other line equations.

Probability of Lines HITTING ABCD = (Sector covered by North and South quadrant) / 360 = (90 x 2) / 360 = 1 / 2.

Probability of Lines NOT HITTING ABCD = 1 - (1 / 2) = 1 / 2.

Answer (C).

I agree that the question is a little ambiguous in that it doesn't provide information on how the line is generated 'randomly' but in my opinion, the intent of the question is quite clear. The line passes through P and the angle of the line with the x axis takes a random value from 0 to 180 counterclockwise (note that it is a line so it extends indefinitely on both ends so the line making an angle of 270 degrees counterclockwise with the x axis is the same as the line making an angle of 90 degrees counterclockwise with the x axis)

Say, if the y intercept is randomly generated, the only acceptable values for the y intercept lie between 6 and -6 but y intercept can take infinite values so the options don't work in that case.
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I took an approach similar to jcaine:

Looking at the figure, you can determine that the triangle formed by DPC has a 90 degree angle at P.

This can be assumed because triangle formed from D, (x = 4) , P is a 45-45-90 triangle. This is certain because the letters fall on distinct points within the graph. A 45-45-90 triangle is also formed by C, (x=8), P. The sides of each triangle formed here are 2-2-x SQRT 2 (45-45-90 triangle)

Using the property of a straight line = 180 degrees, you can deduct that triangle DPC has a 90 degree angle at P and any intersecting line would fall within 90 degrees out of 180.

Double that to get 180 / 360 degrees = 1/2.

Think of it as a circle and the "range" (degree) that the line can and can not pass through.

I hope this explanation helped.

The extremes that cross C and D are in fact perpendicular therefore their intersection make a 90 degree angle.

Therefore 90/180 = 1/2

Answer is C

Hope it helps
Cheers
J

Now I see what the Q meant to test here...Super horrible grammar..

Hi Karishma,

why are the four dotted areas equal probability?
How did you come about to drawing them?

Point D coordinates (4, 2)
Point C coordinates (8, 2)
So if you extend AD and BC to touch the x axis, you get a rectangle (shown in my figure). Split it into two equal rectangles by dropping a perpendicular from P to DC. Now draw the diagonals of both rectangles as shown. Note that the diagonal of a rectangle divides the rectangle into two equal areas. Since the two rectangles were of equal areas, all four areas will be equal. Basically , you divide the entire region around point P into 4 equal areas - in 2 of them, the line will pass through the square, in 2, it will not.
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let O be the point on CD at right angle from P
apply pyth. thm. on triangles POC and POD.. Now we can notice that PC = PD
hence PCD is a isosceles triangle
therefore Angle CPD = 90
So if 0<= Angle CPD <=90 then line will cross the rectangle
So favorable outcomes are 90 and possible outcomes are 180

so probability will be 90/180 = 1/2
probability that line will not cross through rectangle will be 1 - 1/2 = 1/2

The thought was this:



Any line passing through the point P which is at 6 on the x axis will pass through one of the 4 equal dotted regions (except x axis). Out of these, lines passing through 2 of the regions are not acceptable (red lines) while those passing through other 2 are acceptable (green lines).
Hence, 1/2 of the region is fine.

Probability that the line does not pass through ABCD is 1/2.[/q

Hi Karishma,

How did you arrive at the 4 lines ( red and green)? Please explain.

Hi Karishma,

I am a bit confused. What is the use of y=mx+b???

Hi

I think y = mx+b is just given to state general equation of a line. (meaning this line has a slope of 'm' and a y intercept of 'b')
Using this equation is not required here, as you can see from the previous solutions.

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