Dots In A Circle
There are 40 dots spaced evenly in a circle. If 4 distinct dots W , X , Y and Z are picked randomly, what is the probability that W X intersects Y Z ?
If the probability can be expressed as b a , where a and b are coprime positive integers, find a + b .
The answer is 4.
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4 solutions
This line of reasoning is very nice in that it does not require us to deal with the specific conditions in the problem. A similar approach can be used to solve this problem .
Without loss of generality, take W as given and number the points clockwise from 0 to 39 starting with W at 0. X , Y and Z are therefore chosen at random from whole numbers 1 to 39. The problem is equivalent to asking whether the point assigned to X is between the numbers assigned to Y and Z . Consider then the labels for X , Y and Z . X will have the least label 3 1 of the time and the greatest label 3 1 of the time. In those cases, W X will not intersect Y Z . The segments will intersect in the remaining 1 − 3 2 = 3 1 of the cases. The solution, therefore is 1 + 3 = 4
Total number of ways of selecting 4 distinct vertices are n(n-1)(n-2)(n-3)=(n!)/(n-4)! .Now we calculate the total number of favourable ways.Consider two intersecting diagonals whose point of intersection is not on the circumference.To any such pair of diagonals we can uniquely associate a quadrilateral that has these diagonals same as its own diagonals.Hence, for every pair of permissible intersecting diagonals, we have a unique quadrilateral and vice versa.Because of this bijection,we conclude that the total number of permissible intersecting diagonals is equal to number of ways of forming a quadrilateral = nC4.After selecting a pair of permissible intersecting diagonals,points W,X,Y and Z can be allotted to the endpoints of the diagonals in 4 ∗ 1 ∗ 2 ∗ 1 = 8 ways.Hence total number of favourable ways = 8 ∗ nC4.We have , given probability as favourable ways divided by total number of ways which comes out to be equal to 1/3.Hence, answer is 1+3=4.
There is a one line solution to this problem.
Hint: It doesn't matter how many dots there are (as long as there are more than 4).
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Pick a dot for W and then permute XYZ ( XYZ, XZY, YXZ, YZX, ZXY and ZYX). Reading clockwise WX intersects YZ only if X is in the middle ( 2 cases YXZ and ZXY)...2/6=1/3
(In your hint you mentioned that you needed MORE than 4... Would 4 be sufficient?)
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Yes, that's the idea. Counting the probability by "subcases" avoids having to do the tedious calculation/counting.
Yes, 4 is sufficient.
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@Calvin Lin – The calculations in Indraneel's solution are not at all tedious and can be done orally.You need to read his solution carefully.
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@Manish Maharaj – My point is that
- His solution could be shortened considerably to get at the crux of the problem
- There wasn't a need to involve n in the calculations.
beautiful problem..i solved it in this way: all possible sets of 4 points out of 40 = choose(4,40), this should be multiplied by 3 which are the possible lines between these 4 point. Those are also the possible cases. You then proceed to calculate the favourable cases 1*choose(4,40) since there is only one case in which the lines( that pass trough these 4 points) intersect themselves. favourable/possible = 1/3
Consider any set of 4 dots on the circle, A, B, C, D.
There are 3 ways that we can pair them up.
There is exactly 1 way that we get an intersection, namely when the pairs are separating each other.
Hence, the probability is 3 1 .
Since this holds for any set of 4 dots, the total probability is 3 1 .