Question

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

a) if the percentage of packages that need to be checked is 1% then what is the probability that hour 10 is the first sample at which X exceeds 1

Ok, so I calculated, P(X = 0) and P( X = 1)

P(X = 0) = 0~Binom(0.1,20) = 0.8179
P(X = 1) = 1~Binom(0.1,20) = 0.1636

So, P(X>1) = 0.0185

Let Y be the number of packages that need to be checked every 10 hours.

I'm looking for P(Y>1) = 1 - (P(Y=1) + P(Y=0))

P(Y=0) = (0.8179)^(10)
P(Y=1) = (0.8179)^9(0.1636)

Answer 1

So I get,

P(Y<1) = 1-(0.134 + 0.02679) = 0.83921

Answer 2

Is this correct? or am I way off?

Answer 3

Well, you're not really factoring in what hour 10 means. So I don't quite get that.

Answer 4

Does hour 10 means that \(10\times 20\) packages have gone through?

Answer 5

yes

Answer 6

but I assume we have to look at each hour as an individual trial

Answer 7

The answer I formulated doesnt really make sense to me though

Answer 8

But it sort of does :\

Answer 9

My reasoning is that the order of successes and failures in a geometric series doesnt matter as long as there are a certain number of successes and a certain number of successes

Answer 10

but I could be wrong

Answer 11

Is says hour 10 is the first sample. That means that the previous hours got \(X=0\), doesn't it?

Answer 12

oh wait, now I see...

Answer 13

It says when it exceeds 1

Answer 14

when X exceeds 1 meats X>1

Answer 15

so each one before got 0 or 1.

Answer 16

means*

Answer 17

this question is confusing me so much :\

Answer 18

Then you want to find: \[
[1-\Pr(X> 1)]^9\Pr(X> 1)
\]

Answer 19

the thing that confuses me is that a X = 1 in any of the trials, but X has to equal 2 or more in trial 10

Answer 20

If we treat each hour as being independent.

Answer 21

true, but cant that result in X ending up larger than 1 in one of the 9 trials because they are all independent

Answer 22

or am I way off maybe I'm not thinking correctly about this it is so hot in my apartment

Answer 23

Consider the simple case where the first hour doesn't exceed and the second one does.

Answer 24

Okay, well it is summer.

Answer 25

True enough I guess complaining about weather is not something someone should do unless it is destroying their life

Answer 26

Your concern is that the formula I gave would allow say....

pass exceed pass pass pass pass pass pass pass

for example?

Answer 27

yes

Answer 28

Maybe I dont have a proper understanding of geometric distribution which is probably the case

Answer 29

I guess it doesnt matter if that is the case

Answer 30

because X>2

Answer 31

or equal

Answer 32

Well, what I would say is that:
pass exceed pass pass pass pass pass pass pass
Has the same probability as:
pass pass pass pass pass pass pass pass exceed
If we think of these events as independent.
And the only way to allow every combination is if we multiplied by \(9 \choose 1\).

Answer 33

Or \(10 \choose 1\) I guess.

Answer 34

So the only real question is whether or not the events are independent.

Answer 35

If they aren't independent, then where does that \(1\%\) come from?

Answer 36

They are independent

Answer 37

I guess what you propose makes sense, I may just need to sleep on it I feel like I'm not understanding why it is the right answer.

Answer 38

Do you remember the binomial distribution formula?

Answer 39

yes,

\[\left(\begin{matrix}n \\ x\end{matrix}\right)(1-p)^x(p)^{n-x}\]

Answer 40

Do you understand how it was derived?

Answer 41

I think I kind of understand it but perhaps I should reread the chapter on binomials, so I guess no.

Answer 42

It's very simple. Here, let's come up with a simple example. Consider you want to roll a die 5 times, and you want to get 1 exactly two times.

Answer 43

So in this case \(p=1/6\), \(n=5\), \(x=2\).

Answer 44

If we did this manually, we would start with: \[
\frac 16 \times \frac 16\times \frac 56\times \frac 56\times \frac 56
\]This means we got 1 on the first two attempts.

Answer 45

Now suppose we get 1 on the last two attempts: \[
\frac 56\times \frac 56\times \frac 56\times \frac 16\times \frac 16
\]Due to commutative property of multiplication, these are equivalent.

Answer 46

So we get the:\[
p^{x}(1-p)^{n-x}
\]part here.
And it just becomes a question of where do we distribute the the successes?

Answer 47

And that number comes from \(n\choose x\).

Answer 48

Does that make sense?

Answer 49

yes that does thanks

Answer 50

You are saying since the probability that it happens at the start is equal to the probability that happens at the end, then somehow we are calculating the probability of either happening.

Answer 51

When in fact we need to put a \(10\choose 1\) in front if we want to calculate that.

Answer 52

In which case, the geometric series becomes a binomial distribution with \(k=1\).

Answer 53

Ok that makes sense :), thank you for straightening this out for me

Answer 54

Now one thing to consider... I think is...

Answer 55

What is the whole Y part about?

Answer 56

Seem to me that we let \(n=20\times 10\)?

Answer 57

Yeah but each trial is independent

Answer 58

But in this case, is it okay if you get 2 in the very first trial.

Answer 59

Y is still a binomial bariable at the end of the day, isn't it?

Answer 60

It is in the geometric chapter questions in my textbook which comes before poisson so I dont think it can be treated as a poisson distribution

Answer 61

The question is basically just saying that in the first 9 hours X=1 for one of the trials and X=0 for the rest of the trials but on the 10th hour X = 2 or more

Answer 62

Poisson is not very helpful here because we don't know what \(\lambda\) is, do we?

Answer 63

so that is where I'm conflicted with your answer wio, because more than one trial in your solution can be X=1, thus X=2 can happen anytime during the 10 trials, but I assume order isnt important so maybe that isnt a factor?

Answer 64

My answer gives you the probability that it happens on the first trial and not in the last 9. If give you the probability it happens on the last trial but not the first 9. It doesn't give you the probability that either of these events happen.

Answer 65

For it to do that, you'd have to multiply by 2.

Answer 66

Wait, do you mean my \(n=10\times 20\) thing?

Answer 67

ugh ok ok I understand ughhhhh I'm being dumb sorry

Answer 68

maybe you are right perhaps you could just take,

1 - (0~biom(0.1,200) + 1~biom(0.1,200) ) = P(Y>1)

Answer 69

Ah, biom distribution.

Answer 70

I should check to see if it gives me the same answer I got 0.0156 for the one way already went over

Answer 71

http://www.wolframalpha.com/input/?i=1+-+%28%281-0.01%29%5E200+%2B+200*%281-0.01%29%5E199%280.01%29%29

Answer 72

got a different answer :\

Answer 73

Why should they give the same answer?

Answer 74

If you checked 2 packages in the first hour, you would not be counted for the first case, but you would for the second case.

Answer 75

ok ok I think I get it thank you so much for your help :)

Answer 76

Using the binomial distribution I calculated the probability that 2 or more packages need checking in an hourly check of 20 to be 0.0184. Then using the geometric distribution the probability that 2 or more packages need checking for the first time at hour10 is:
\[p(10)=0.0184\times0.9816^{10-1}=0.0156\]

Answer 77

that is also what I got Kropot72

Answer 78

Yes, I had noticed that and though it was worth posting. Sorry about posting so many misguided attempts at this (I have deleted them).