What's the probability ?

Question

Given a circumference of radius r, we choose four points ( a,b,c, and d) around it randomly (one after another in that order). Then we draw two chords connecting point a with point b, and point c with point d.
So, what's the probability that both chords intersect themselves ?
Please explain the solution.

Answer 1

Draw a circle.
Draw point A on the circle.
Draw three other points on the circle, but don't label them.

Two of the unlabeled points are adjacent to A. The third is not.
The chords intersect if and only if B is the point not adjacent to A.
There's a 1/3 chance that this point is the point B.
Therefore, the probability that the chords intersect is 1/3.

Answer 2

i think my argument will no longer be nicely won. I say the risk is 50% permit a ??, permit b ?? and randomly opt for the values for a and b. As already mentioned, for a ? 0, P( a < b²) = a million, that's trivial. purely somewhat much less trivial is the theory that P(a < 0 ) = a million/2 and hence P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what happens while a > 0 For a > 0, whilst this is elementary to tutor there's a non 0 risk for a finite b, the cut back, the risk is 0. a < b² is reminiscent of exclaiming 0 < a < b², remember we are purely finding at a > 0. If this a finite era on an unlimited line. The risk that a is an component of this era is 0. P( a < b² | a > 0) = 0 As such we've a entire risk P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 remember, this is because of the countless instruments. no remember what form of era you draw on paper or on a working laptop or computing gadget you will detect a finite risk that seems to attitude a million. yet that's because of the finite random huge form turbines on the computing gadget and if we had this question asked with finite values there may well be a a answer greater suitable than 50%. i do no longer propose to be condescending, yet please clarify why utilising the Gaussian to approximate a uniform distribution is a stable theory? are not countless numbers relaxing. Cantor while mad working with them! :)

Answer 3

First of all chords cant 'intersect themself'

You mean if the chords intersect eachother.

1/3

Fix a anywhere.

Then starting to count clockwise from a we have 3! = 6 arrangements for the other 3 (b,c,d)

They are

abcd
abdc
acbd
acdb
adbc
adcb

Of which they intersect when b is not adjecant to a.

Clearly there are 2 situations where b and a are not adjecant (acbd) and (adbc)

So they intersect with probability 1/3

and dont intersect with probability 2/3

This is assuming there by 'random' you imply a continuous density at least and the distributions are identical and independant

EDIT

LINYLONS you are wrong.

Chords lie on the edge not inside... the other question IS NOT the same as this, the other question clearly stated 'inside the circle' and is more complex

Answer 4

As long as the four points are not allowed to "pass" each other or extend longer than once around the circumference, then the probability is zero! Draw it and you'll see. One chord would have to begin before the previous one ended for them to intersect and it sounds like the rules don't allow that.

Answer 5

If the order is abcd (in, say, clockwise order starting at (1,0) ) they don't intersect with each other. More generally, if a and b are neighbors in the cyclical order.

If the order is acbd (more generally, if a and b are not neighbors in the cyclical order) they do intersect.

You may like to make a drawing to illustrate this.

a has two neighbors in the cyclical order: b+c, b+d or c+d. Only one of the three possibilities leads to intersection. So the probability is 1/3.

Answer 6

1/6 let abcd be a quadrilateral it can be joined by six ways viz., the four sides and two diagonals. but they intersect in only one way i.e the diagonals which is only one .