Question

20 packages are processed through customs each hour, let X denote the number of packages that need to be checked every hour.

If the rework percentage increases to 4%, what is the expected number of hours until X exceeds 1?

SO, I have the function,

(1-0.1897)^(n-1)(0.1897) = 1

where, 0.1897 = P(X>1)

I get a negative value for n what am I doing wrong?

Answer 1

or this is a property of the geometric distribution :\

Answer 2

expected number... this implies we are dealing with expected value.

Answer 3

Lack of memory property is responsible but then how do I get figure out how many trials until I have a guaranteed success?

Answer 4

You can never guarantee success. Expected value is the average.

Answer 5

It's possible they go 1 billion years without checking a bag, though the probability is very, very low.

Answer 6

\[
E[X] = \sum_i x_i\times \Pr(X=x_i)
\]

Answer 7

so does this question have no answer then?

Answer 8

Or rather there is no number of trials that guarantee sucess

Answer 9

No...

Answer 10

You need to calculate: \[
\sum_i \color{red}{x_i}\color{blue}{\Pr(X=x_i)}= \sum_{n=0}^{\infty} \color{red}{n}\color{blue}{(1-p)^{n-1}p}
\]

Answer 11

Which essentially will derive the mean of the geometric series, which you probably could just as easily look up.

Answer 12

The mean of a geometric series is simply 1/p

Answer 13

how can I use that to figure out how many trials need to be conducted though

Answer 14

The mean IS the expected value.

Answer 15

Oh yeah... ugh

Answer 16

sorry for being so slow

Answer 17

So did you get it then?