K*V
发帖数: 192 | 1 Let x, y be uniformly distributed on [0,1] and separate the unit interval [0
,1] into 3 pieces, what is the probability that the 3 pieces of the line can
be constructed into a right-angled triangle? | K*V
发帖数: 192 | 2 如果只是一般三角形还好，但是right-angled就不会了。
[0
can
【在 K*V 的大作中提到】
: Let x, y be uniformly distributed on [0,1] and separate the unit interval [0
: ,1] into 3 pieces, what is the probability that the 3 pieces of the line can
: be constructed into a right-angled triangle?
| k*****n
发帖数: 117 | 3 0
【在 K*V 的大作中提到】
: 如果只是一般三角形还好，但是right-angled就不会了。
:
: [0
: can
| K*V
发帖数: 192 | 4 别闹了！
【在 k*****n 的大作中提到】
: 0
| k*****n
发帖数: 117 | 5 Did not!
>>> f = lambda (x,y): min( abs(np.array([ x*x, (y-x)*(y-x), (1-y)*(1-y) ])*2
.0 - x*x-(y-x)*(y-x)-(1-y)*(1-y)) ) < 1e-9
>>> data = np.random.rand(1000000,2)
>>> sum( map( f, data ) )
0
Here's why:
for each x, there is at most 2 possible y value to form right triangle.
in x-y plane, your sample space is the unit square (0,0) to (1,1)
but your outcome space is only two 1-d curves.
The probability is 0.
【在 K*V 的大作中提到】
: 别闹了！
| K*V
发帖数: 192 | 6 hehe.It seems make sense...........
Thanks!
*2
【在 k*****n 的大作中提到】
: Did not!
: >>> f = lambda (x,y): min( abs(np.array([ x*x, (y-x)*(y-x), (1-y)*(1-y) ])*2
: .0 - x*x-(y-x)*(y-x)-(1-y)*(1-y)) ) < 1e-9
: >>> data = np.random.rand(1000000,2)
: >>> sum( map( f, data ) )
: 0
: Here's why:
: for each x, there is at most 2 possible y value to form right triangle.
: in x-y plane, your sample space is the unit square (0,0) to (1,1)
: but your outcome space is only two 1-d curves.
| r*******t
发帖数: 8550 | 7 veli smart!
【在 k*****n 的大作中提到】
: 0
| t********t
发帖数: 1264 | 8 别闹，拿measure theory 101来当面试题，羞辱人啊 |
|
