# I got this problem in my book.. I will directly go the last part which I could not solve. P(good condition)=0.91854, P(bad condition)=0.08146. A 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).

Geometric distribution and Conditional probability problem
I got this problem in my book.. I will directly go the last part which I could not solve.
$P\left(goodcondition\right)=0.91854$
$P\left(badcondition\right)=0.08146$
a 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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tal1ell0s
Step 1
$P\left(X>5\right)=1-P\left(X\le 5\right)={0.91854}^{5}\approx 0.6539$
To calculate it you can use the fact that the CDF of your rv is known...
Step 2
If 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 $0.91854×0.91854×0.91854×0.91854×0.91854×1\approx 0.654$