Geometric probability - parallelogramsInside a rhombus E with sides 10 unit and one interior angle less than 90 degree, there are 2 parallel (with E) parallelograms A and B, both can move freely and uniformly inside E but must keep parallel with E in moving. A is with base 8 unit and adjacent side 6 unit; while B is with base 5 unit and adjacent side 9 unit. If a point is chosen randomly in E, find the probability that the point lies inside A and B at the same time.

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Izabella Fisher · Accepted Answer

Step 1We can either impose oblique (x,y) coordinates on the plane of the three paralellograms, with the x- and y-axes parallel to the sides of all the parallelograms, or we can transform the plane linearly (preserving relative areas) so that the rhombus becomes a square. Either way, it is convenient to place the axes so that the bottom and left sides of the rhombus lie along the x- and y-axes, respectively.Taking the random variables       X    A    ,      Y    A  ,       X    A    ,      Y    A   as the coordinates of the lower left corner of parallelogram A,       X    B    ,      Y    B   as the coordinates of the lower left corner of parallelogram B, and       X    P    ,      Y    P   as the coordinates of a random point P within the rhombus, let the variables have pairwise independent distributions                              X          A                                    ∼        U        (        0        ,        2        )                                      Y          A                                    ∼        U        (        0        ,        4        )                                      X          B                                    ∼        U        (        0        ,        5        )                                      Y          B                                    ∼        U        (        0        ,        1        )                                      X          P                                    ∼        U        (        0        ,        10        )                                      Y          P                                    ∼        U        (        0        ,        10        )            where U (a, b) is the uniform distribution on the interval [a,b].Step 2Now, P is (strictly) inside   A  ∩  B if and only if                     (Q)                    min        {                  X          A                ,                  X          B                }        &amp;lt;                  X          P                &amp;lt;        max        {                  X          A                +        8        ,                  X          B                +        5        }            and                     (R)                    min        {                  Y          A                ,                  Y          B                }        &amp;lt;                  Y          P                &amp;lt;        max        {                  Y          A                +        6        ,                  Y          B                +        1        }        .            (If a point on the boundary of   A  ∩  B counts as &quot;inside both A and B&quot; then change &amp;lt; to ≤ in the previous statement.)Since the variables are pairwise independent, so are the events Q and R described by equations (Q) and (R), respectively, and the answer to the question,   P  (  Q  ∩  R  ), obeys the equation   P  (  Q  ∩  R  )  =  P  (  Q  )  P  (  R  )  ..So consider equation Q. For given values       X    A    =      x    A   and       X    B    =      x    B  , let   u  =  min  {      X    A    ,      X    B    } and   v  =  max  {      X    A    +  8  ,      X    B    +  5  }; then  P  (  Q  ∣      X    A    =      x    A    ,      X    B    =      x    B    )  =  P  (  u  &amp;lt;      X    P    &amp;lt;  v  )  =      1    10    (  v  −  u  )(since   0  ≤  u  ≤  v  ≤  10). So the unconditional probability P(Q) is just       1    10   of the mean size of the interval   [  min  {      X    A    ,      X    B    }  ,  max  {      X    A    +  8  ,      X    B    +  5  }  ]. I figure the mean of   max  {      X    A    +  8  ,      X    B    +  5  }  −  min  {      X    A    ,      X    B    } is       71    15  , so   P  (  Q  )  =      71    150  , and by similar reasoning   P  (  R  )  =      71    120  , so I find that  P  (  Q  ∩  R  )  =      (          71      150        )        (          71      120        )    =  5041      /    18000  ≈  0.2800555  …  .

Dillan Valenzuela · Accepted Answer

Step 1We may assume that everything happens in the unit square   [  0  ,  1      ]    2  .Begin with the following one-dimensional problem: We have a movable subinterval   J  ⊂  [  0  ,  1  ] leaving free space of length   0  &amp;lt;  ℓ  ≤            1      2      , whose position is uniformly distributed within the given limits. Denote by p(x) the probability that the point   x  ∈  [  0  ,  1  ] is covered by J. Then   p  (  1  −  x  )  =  p  (  x  ), by symmetry. For   0  ≤  x  ≤            1      2       it is easy to see that   p  (  x  )  =      {                                                                                        x                                                                                                              (                                                                                      ℓ                                                                                                                                    (              0              ≤              x              ≤              ℓ              )                            ,                                                            1                                                                                      (                                            ℓ              ≤              x              ≤                                                1                  2                                                                              )                                                          .                                              .Step 2If we now have two such intervals       J    1  ,       J    2   such that       ℓ    1    &amp;lt;      ℓ    2    ≤            1      2      , distributed independently, then the probability q(x) that x is covered by both intervals computes to  q  (  x  )  =      {                                                                                                            x                    2                                                                                                                                (                                                                                                            ℓ                      1                                                              ℓ                      2                                                                                                                                                        (              0              ≤              x              ≤                              ℓ                1                            )                            ,                                                                                                                                                                                                                    (                                                                                      x                                                                                                                                (                                                                                                            ℓ                      2                                                                                                                                                        (                              ℓ                1                            ≤              x              ≤                              ℓ                2                            )                            ,                                                            1                                                                                      (                                                            ℓ                2                            ≤              x              ≤                                                1                  2                                                                              )                                                          .                                              Step 3In the case at hand we have       ℓ    1    =            1      5      ,       ℓ    2    =            1      2       for the x-direction. If we now assume that x is uniformly distributed in [0,1] as well the overall probability       P    x   that the random point x is covered by both intervals becomes      P    x    =  2      ∫    0          1              /            2        p  (  x  )    d  x  =  2      ∫    0          1              /            5        10      x    2      d  x  +  2      ∫          1              /            5              1              /            2        2  x    d  x  =            71      150           .Similarly for the y-direction: Here       ℓ    1    =            1      10      ,       ℓ    2    =            2      5      . The overall probability       P    y   that a random point y is covered by both intervals becomes      P    y    =  2      ∫    0          1              /            10        25      x    2      d  x  +  2      ∫          1              /            10              2              /            5                          5        x            2          d  x  +  2      ∫          2              /            5              1              /            2        1    d  x  =            71      120           .The probability P that a uniformly distributed random point   (  x  ,  y  )  ∈  [  0  ,  1      ]    2   is covered by both random rectangles at the same time is therefore given by  P  =      P    x    ⋅      P    y    =            5041              18                000              ≐  0.280056     ,.

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