Cumulative geometric probability (greater than a value)Emelia registers vehicles for the Department of Transportation. Sports utility vehicles (SUVs) make up 12% of the vehicles she registers. Let V be the number of vehicles Emelia registers in a day until she first registers an SUV. Assume the type of each vehicle is independent. Find the probability that Emelia registers more than 4 vehicles before she registers an SUV.

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decoratesuw · Accepted Answer

Step 1Each vehicle is selected at random, with probability of selecting an SUV being 0.12, since 12% of cars are SUVs. The geometric distribution gives the number of trials of &quot;SUV or not SUV?&quot;, as you examine each car in turn, before the first SUV is found, so this occurs with probability   P  (  k   trials before first SUV  )  =  (  1  −  p      )          k      −      1        p  =      0.88          k      −      1        ×  0.12.In fact, it is better to use the &quot;CDF of the geometric distribution&quot;, which says   P  (            k         trials or fewer before the first success    )  =  1  −  (  1  −  p      )    k  .So, the probability of 4 trials or fewer before the first SUV is given by   P  (  4   trials or fewer before the first SUV  )  =  1  −  (  1  −  p      )          4        =  1  −  (  1  −  0.12      )          4        =  0.40so more than 4 vehicles is   1  −  0.40  =  0.60.Step 2The fact you mention P(MORE than 4 vehicles BEFORE she registers an SUV) is equal to P(FIRST 4 cars not SUVs) is the idea of reversing the probability, so more than 4 is the negation of the event less than or equal to 4, so just do   P  (  event occurs  )  =  1  −  P  (  event does not occur  ) which is why you go from 0.4 to   1  −  0.4  =  0.6 after using the CDF.

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