Probability, geometric distributionWhat is the probability of P ( X &lt; E ( X ) ) if X has a geometric distribution such that P ( X = 1 ) = 0.25.

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Probability, geometric distributionWhat is the probability of   P  (  X  &amp;lt;  E  (  X  )  ) if X has a geometric distribution such that   P  (  X  =  1  )  =  0.25.

Cagliusov8 · Accepted Answer

Step 1We have that   Pr  {  X  =  k  }  =  (  1  −  p      )          k      −      1        p for   k  ≥  1 and   Pr  {  X  =  1  }  =  (  1  −  p      )    0    p  =  p  =  0.25 so that   E  ⁡  X  =      1    p    =  4..Step 2We obtain that   Pr  {  X  &amp;lt;  E  ⁡  X  }  =  Pr  {  X  &amp;lt;  4  }  =      ∑          k      =      1        3    Pr  {  X  =  k  }  =      ∑          k      =      1        3    (  0.75      )          k      −      1        0.25  ≈  0.5781.

What is the probability of P(X<E(X)) if X has a geometric distribution such that P(X=1)=0.25.

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