Probability of geometric mean being higher than 1/6If we choose 2 numbers from [0,2], independent from each other,what is the probability of their geometric mean being higher than 1/6?

Question

Kyle George · Accepted Answer

Step 1Note that       X    Y    &amp;gt;  1      /    6 iff   Y  &amp;gt;  1      /    (  36  X  ) where X, Y are i.i.d uniform [0,2] random variables. Since X,Y are i.i.d uniform [0,2] it follows that   P  (  (  X  ,  Y  )  ∈  A  )  =            m      (      A      ∩      [      0      ,      2              ]        2            )        4   i.e. (X,Y) is uniformly distributed on   [  0  ,  2      ]    2   (here m means area) and   A  ⊂            R        2  ). Letting   A  =  {  (  x  ,  y  )  ∈            R      2        ∣  36  x  y  &amp;gt;  1  } we have that   P  (      X    Y    &amp;gt;  1      /    6  )  =  P  (  Y  &amp;gt;  1      /    (  36  X  )  )  =      1    4        ∫          0        2        ∫          1              /            36      x        2      d  y    d  x which you can compute.

janine83fz · Accepted Answer

Explanation:The values of x and y, you are interested in, lie in a square   [  0  ;  2  ]  ×  [  0  ;  2  ]. So you have a chart of   y  =      1          36      x      , and an answer to your problem equals to an area of intersection of an epigraph and a given square divided by an area of a square. Which equals to             4      −              ∫                  1                      /                    72                          2                                      1                      36            x                              −              2        72              4

If we choose 2 numbers from [0,2], independent from each other,what is the probability of their geometric mean being higher than 1/6?

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