Geometric distribution and the probability of getting success at "first try"?If one knows the probability of success p,Then how does one calculate the probability of getting success at "first try" using geometric distribution?Is it simply the probability of success? Are the successive events independent?

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Geometric distribution and the probability of getting success at &quot;first try&quot;?If one knows the probability of success p,Then how does one calculate the probability of getting success at &quot;first try&quot; using geometric distribution?Is it simply the probability of success? Are the successive events independent?

Julianna Bell · Accepted Answer

Step 1If the random variable has a geometric distribution, then it is the count of trials until the first success in an indefinite sequence of independent Bernoulli trials with an identical success rate. [Sometimes the count of failures before the first success, depending on text. We&#039;ll assume the former.]If   X  ∼      G    e          o      1        (  p  ) then       P    (  X  =  k  )  =  p  (  1  −  p      )          k      −      1                    [        k  ∈  {  1  ,  .  .  ,  ∞  }            ]      .Step 2You appear to require       P    (  X  =  1  ), the probability that there is one trial until the first success. (The count of trials until the first success is one.)

Geometric distribution and the probability of getting success at "first try"?

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