Why is P ( X &gt; r ) = q r ?If X follows a geometric distribution, where p = probability of success and q = probability of failure, why is P ( X &gt; r ) = q r ?

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Why is   P  (  X  &amp;gt;  r  )  =      q    r  ?If X follows a geometric distribution, where   p  = probability of success and   q  = probability of failure, why is   P  (  X  &amp;gt;  r  )  =      q    r    ?

stangeix · Accepted Answer

Step 1Because it means that you have failed for at least the first r times.Since the only sequence that has this outcome is the one when you always fail, and you fail r times with probability q, you have   P  (  X  &amp;gt;  r  )  =      q    r  .Step 2Let&#039;s compute   P  (  X  =  r  +  1  )  +  ⋯  +  P  (  X  =  n  )We have   p      q          r        +  ⋯  +  p      q          n      −      1        =  p      q    r    (  1  +  ⋯  +      q          n      −      r      −      1        )  =  (  1  −  q  )      q    r                      q                  n          −          r                    −      1              q      −      1        =      q    r    (  1  −      q          n      −      r        )  =      q    r    −      q          n      Now letting   n  →  ∞ you can see that       q    n    →  0, so   P  (  X  &amp;gt;  r  )  =      lim          n      →      ∞        P  (  X  =  r  +  1  )  +  ⋯  +  P  (  X  =  n  )  =      lim          n      →      ∞            q    r    −      q    n    =      q    r

If X follows a geometric distribution, where p= probability of success and q= probability of failure, why is P(X>r)=qr?

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