A problem on geometrical probabilityTwo points P and Q are taken at random on a straight line OA of length a,show that the chance that P Q &gt; b, where b &lt; a is ( a − b a ) 2

Question

A problem on geometrical probabilityTwo points P and Q are taken at random on a straight line OA of length a,show that the chance that   P  Q  &amp;gt;  b, where   b  &amp;lt;  a is   (                    a        −        b            a            )    2

Gauge Howard · Accepted Answer

Step 1You are looking for the probability   P  (  Z  &amp;gt;  b  )  =  P  (      |    X  −  Y      |    &amp;gt;  b  ) where X and Y are uniform and independent random variables on [0,a]. In case the random variables are independent, their sum corresponds to the convolution of their density functions. I hope you know what a probability density function and convolution are.For example if we have the density of Z, then easily   P  (  Z  &amp;gt;  b  )  =      ∫          Z      &amp;gt;      b            p    Z    (  z  )  d  zIn this example it is not necessary to calculate the density function of Z. It is enough to calculate the density of   X  −  Y. If we let   W  =  −  Y, then we have   X  +  W. What is the density of W? it is uniform on [-a,0].if you calcluate       p          X      +      W        (  x  )  =      ∫          −      ∞        ∞        p    X    (  y  )      p    W    (  x  −  y  )  d  y  =      ∫          −      ∞        ∞        p    X    (  x  −  y  )      p    W    (  y  )  d  yyou will obtain       p          X      +      W        (  x  )  =            a      −      x              a      2        (  u  (  x  )  −  u  (  x  −  a  )  )  +            x      +      a              a      2        (  u  (  −  x  )  −  u  (  −  x  −  a  )  ) where u is the unit step function.Step 2Now what does       |    X  −  Y      |    &amp;gt;  b mean? it means   X  −  Y  &amp;gt;  b if   X  &amp;gt;  Y AND   Y  −  X  &amp;gt;  b if   Y  &amp;gt;  X. For the last case if you multiply both sides by -1, you will get   X  −  Y  &amp;lt;  −  bThis means we need to integrate the function       p          X      +      W       over two regions   x  &amp;lt;  −  b and   x  &amp;gt;  b. That is   P  (      |    X  −  Y      |    &amp;gt;  b  )  =      ∫          −      a              −      b            p          X      +      W        (  x  )  d  x  +      ∫          b        a        p          X      +      W        (  x  )  d  x.The result of both integrals are   A  =      1    2              a      −      b              a      2        (  a  −  b  ) and hence   P  (      |    X  −  Y      |    &amp;gt;  b  )  =  2  A  =            (      a      −      b              )        2                    a      2      .Note that the term   h  =  (  a  −  b  )      /        a    2   is the heigth of the little red triangle in zap&#039;s answer and   w  =  (  a  −  b  ) is its width. The triangle function is       p          X      +      W      .

Maghrabimh · Accepted Answer

Step 1Probably not the most elegant solution, but we can use calculus.Assign the leftmost and rightmost point of OA to 0 and a respectively, and x and y be the coordinate of points P and Q. We have Step 2  P  (      |    y  −  x      |    &amp;gt;  b  )  =                    ∫        0                  a          −          b                            ∫                  x          +          b                a            d      y             d      x                      ∫        0        a                    ∫        x        a            d      y             d      x        =            (                  a      −      b        a                      )              2

Two points P and Q are taken at random on a straight line OA of length a,show that the chance that PQ>b,where b<a is ((a-b)/a)^2

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