Geometric distribution and Conditional probability problemI got this problem in my book.. I will directly go the last part which I could not solve. P ( g o o d c o n d i t i o n ) = 0.91854 P ( b a d c o n d i t i o n ) = 0.08146a random package check is checked by a company till there is a package in a bad condition .1. What is the probability that the company will check exactly 4 packages?2. It is known that the first 3 packages that were checked are in good condition , what is the probability that more than 8 packages will be checked(edit: the answer should be 0.654).

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Geometric distribution and Conditional probability problemI got this problem in my book.. I will directly go the last part which I could not solve.  P  (  g  o  o  d  c  o  n  d  i  t  i  o  n  )  =  0.91854  P  (  b  a  d  c  o  n  d  i  t  i  o  n  )  =  0.08146a random package check is checked by a company till there is a package in a bad condition .1. What is the probability that the company will check exactly 4 packages?2. It is known that the first 3 packages that were checked are in good condition , what is the probability that more than 8 packages will be checked(edit: the answer should be 0.654).

tal1ell0s · Accepted Answer

Step 1  P  (  X  &amp;gt;  5  )  =  1  −  P  (  X  ≤  5  )  =      0.91854    5    ≈  0.6539To calculate it you can use the fact that the CDF of your rv is known...Step 2If you do not know the CDF of a geometric distribution you can do the following reasoning...the probability to have more than 5 failures is exactly the probability of having 5 consecutive failures...after this events any event can happen....thus you probability is   0.91854  ×  0.91854  ×  0.91854  ×  0.91854  ×  0.91854  ×  1  ≈  0.654

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