# Poisson process

Hi,
I have This probability problem and I dont know how to do it:

A company makes plastic panel used in automobiles. The panel production process is such thast the number of flaws on a panel follows a Poisson process with a mean of 0.03 flaws per panel.

1- If one panel is randomly selected from the production process, what is the probability it has no flaws.
My solution:
lambda=0.03*1=0.03
f(0,0.03)=(e^(-0.03)*(0.03)^0)/(0!)

2- if 50 panesl is randomly sampled from the production process, what is the probability it has no flaws.
No solution here. I am not sure but it is the same as above but with lambda=0.03*50??

3- What is the expected number of panels that need to be sampled before flaws are found?
No solution indeed!!

Please can I have some help with this problem?
B.

## The Attempt at a Solution

2. Yes, you are right. lambda can be replaced with (n*p), where n is the # of trials, and p is the probability. Since you increase your trials to 50, np = 50*p.

3. expected number is summation(x*Poisson). Also, you can think of expected number as the number of trials you need to 'expect' your first flaw. Since you have a 3% chance of expecting a flaw in any given panel, independent of any other panel, you can 'expect' to see a flaw on your 34th panel you sample....yea?

thank you,
But what is, in the expected number formula ,the value of lambda?(if we want to use the formula)

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lambda can be anything, as long as you define it correctly. Let's use your lambda = .03, where lambda is the number of flaws in any one panel.

If you do the expected number summation I gave you above, I think you will get exactly 'lambda' as your answer. Since you want to know when there will be '1' flaw (and not .03 flaws), you will have to multiply .03 by a number 'n', until you get (.03*n) = 1. 'n' happens to be 33.3333..., which will round up to 34 panels, before you see 1 flaw. Hope that answer works.

OK
thank you mkkrnfoo85!

I have another problem and would like to have some help please.

A company rents time on a computer for periods of t hours, for which it receives \$600 an hour. The number of times the computer breaks down during t hours is a random variable having the Poisson distribution with lambda=(.8)t, and if the computer breaks down x times during t hours, it costs 50x^2 dollars to fix it.

-How should the company select t in order to maximize its expected profit?
I dont know how to compute the expected profit here so that I can use the equation to find t.

Do you have any idea?

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