Finding Geometric Probability?Four points are chosen at random from the interior of a circle. What is the probability that the sum of each of their distances from the center is greater than half the circumference of the circle?

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Allvin03 · Accepted Answer

Step 1Assume the disk have radius 1.If you have   r  ∼  U  n  i  [  0  ,  1  ], i.e. random radius and   θ  ∼  U  n  i  [  0  ,  2  π  ], i.e. random angle, then   X  =      r    cos  ⁡  θ and   Y  =      r    sin  ⁡  θ have a joint pdf on the disk       1    π  , i.e. describes a point picked uniformly at random on the disk.Step 2And   R  =            X      2        +          Y      2

Jamarcus Lindsey · Accepted Answer

Step 1Denote by R the absolute value of a uniformly distributed random point Z in the unit disc. The cdf and the pdf of R are then given by                    (1)                              F          1                (        s        )        =                  {                                                    0                                                            (                s                &amp;lt;                0                )                                                                                      s                  2                                                                            (                0                ≤                s                ≤                1                )                                                                    1                                                            (                s                ≥                1                )                                                                       ,                          f          R                (        s        )        =                  {                                                    0                                                            (                s                &amp;lt;                0                )                                                                    2                s                                                            (                0                &amp;lt;                s                &amp;lt;                1                )                                                                    0                                                            (                s                &amp;gt;                1                )                                                                       .            The index 1 refers to the first random point. Using (1) one computes the cdf       F    2    (  s  ) of       R    1    +      R    2   for two random points, and then       F    3    (  s  ),       F    4    (  s  ) as follows:      F    2    (  s  )  =      ∫    0    1        f    R    (  t  )      F    1    (  s  −  t  )    d  t  ,    …  ,        F    4    (  s  )  =      ∫    0    1        f    R    (  t  )      F    3    (  s  −  t  )    d  t     .Step 2Mathematica obtained the following expression for       F    4    (  s  ) in the interval   3  &amp;lt;  s  &amp;lt;  4:                    31192        −        12288        s        −        17920                  s          2                +        12544                  s          3                −        1680                  s          4                −        448                  s          5                +        112                  s          6                −                  s          8                    2520           .The probability p that       R    1    +      R    2    +      R    3    +      R    4    ≥  π then comes to   p  =  1  −      F    4    (  π  )  ≈  0.1625     .

Four points are chosen at random from the interior of a circle. What is the probability that the sum of each of their distances from the center is greater than half the circumference of the circle?

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