On the interval [0, 5] we randomly and independently choose two numbers that divide this interval into three sections. The smaller number is denoted by a and the larger number by b. What is the probability event that section [a, b] will be the shortest and section [0, a] the longest?

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spornya1 · Accepted Answer

Step 1Let x and y be the two numbers, with   a  =  m  a  x  (  x  ,  y  ),   b  =  m  i  n  (  x  ,  y  ).Let E be the event of interest.Notice first that   P  (  E      |    x  &amp;gt;  y  )  =  P  (  E      |    x  &amp;lt;  y  ), by symmetry, hence   P  (  E  )  =  P  (  E      |    x  &amp;gt;  y  ). Let&#039;s compute   P  (  E      |    x  &amp;gt;  y  ); then, we assume that   x  &amp;gt;  y, hence   b  =  x and   a  =  y.Step 2The conditions on the three lengths can be written as   b  −  a  &amp;lt;  5  −  b  &amp;lt;  a.That is,   y  &amp;gt;  2  x  −  5 and   y  &amp;gt;  5  −  xThose inequalities corresponds to a region on the   x  −  y plane. Because the condition dictates that the points are uniform inside a triangle   0  &amp;lt;  y  &amp;lt;  x  &amp;lt;  5, we only need to intersect that with the above requirements, and compute the relative area.

On the interval [0, 5] we randomly and independently choose two numbers that divide this interval into three sections. The smaller number is denoted by a and the larger number by b. What is the probability event that section [a, b] will be the shortest and section [0, a] the longest?

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