Let X ∼ G e o m ( 0.75 ). Find the probability that X is divisible by 3

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Let   X  ∼  G  e  o  m  (  0.75  ). Find the probability that X is divisible by 3

unfideneigreewl · Accepted Answer

Step 1There are two versions of the geometric distribution. In one of them, the random variable X measures the number of trials until the first success, including the trial that gave the success. In the other version, the random variable Y counts the number of failures until the first success.In elementary courses, the first version is more common than the second. So we use that one. Thus the possible values of X are 1,2,3,4,….So X is divisible by 3 if the number of trials is 3,6,9,12,….Step 2We have   Pr  (  X  =  3  )  =  (  0.25      )    2    (  0.75  ) (2 failures and then success). Similarly,   Pr  (  X  =  6  )  =  (  0.25      )    5    (  0.75  ), and   Pr  (  X  =  9  )  =  (  0.25      )    8    (  0.75  ), and so on. Thus the probability that X is divisible by 3 is   (  0.25      )    2    (  0.75  )  +  (  0.25      )    5    (  0.75  )  +  (  0.25      )    8    (  0.75  )  +  (  0.25      )          11        (  0.75  )  +  ⋯  .This is an infinite geometric series, with first term   a  =  (  0.25      )    2    (  0.75  ) and common ratio   r  =  (  0.25      )    3  . By a standard formula, the sum of the geometric series is equal to       a          1      −      r        .

Let X~ Geom(0.75). Find the probability that X is divisible by 3

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