View Full Version : A probability question

tennis89

07-09-2010, 02:17 PM

a stick of length 1 is cut twice randomly, probability that 3 segments can form a triangle?

x,y, 1-x-y

x+y>1-x-y ==> x+y>1/2

1-x-y+x>y ==> y<1/2

1-x-y+y>x ==> x<1/2

we can find the area in the x-y coordinate (within the square composed of (0,0),(0,1),(1,0),(1,1)), is 1/8, so I think the probability is 1/8 but the answer is 1/4, did I miss something here?

Thanks

Dason

07-09-2010, 02:27 PM

How is it "randomly cut". I'm assuming a uniform distribution across the length of the stick but I guess my question is how the second cut is chosen. Do they just uniformly pick 2 spots and cut those or do they choose a spot, make the cut, and then choose a spot on the stick to the right. I'm guessing it's the first option but I just want to make sure.

tennis89

07-09-2010, 02:41 PM

both cut are uniform distribution

Mean Joe

07-10-2010, 05:29 PM

Your possible area is NOT in the whole square, eg x=1/2 y=1 is not possible.

yurdakul

07-15-2010, 10:58 PM

You need to consider your sample space as {(x,y)|x+y<1} which has area 1/2 then compare it with your area which is 1/8 then you will get the right answer that is 1/4.

Yes. Actually you need some sort of ordering after randomly generating the 2

independent uniform cutting location.

Either you have ordered the piece after the cut so that

x is the length of the left piece (distance between the left cutting point

to the left end point), and y is the length of the right piece

(distance between the right cutting point to the right end point)

and thus x + y < \frac {1} {2} as those 2 pieces will not overlap.

Or you pick the smaller one to be x, larger one to be y (or vice versa).

Both are the distance from the cutting point to the left end point.

Then the length of the three pieces should be

\{x, 1 - y, y - x \}

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