Poisson approximation distribution

Yes, X' and Y' are also independent. This is because the complement of an independent event does not affect the probability of the other event occurring.
  • #1
drawar
132
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Homework Statement


In a manufacturing process for electrical components, the probability of a finished component being defective is 0.012, independently of all others. Finished components are packed in boxes of 100. A box is acceptable if it contains not more than 1 defective component. Justify the use of a Poisson approximation for the distribution of the number of defective components in a box. 8 boxes are randomly chosen. Given that all of them are acceptable, estimate the conditional probability that they contain exactly 6 defective components altogether.

Homework Equations





The Attempt at a Solution


Let X be the number of defective components in a box containing 100 components, then X~B(100,0.012) => X~P(1.2)
P(X<=1) = 0.663
Let Y be the number of acceptable boxes in 8 randomly chosen ones, thus Y~B(8,0.663)
P(Y=8) = 0.663^8 = 0.0372
P(required) = P(8 acceptable boxes contain 6 defective components) / P(8 acceptable boxes)
I could calculate the probability that 8 randomly chosen boxes contain 6 defective components (using a Poisson distribution with mean=9.6, I guess), but I'm clueless when it comes to '8 acceptable boxes'.
Any feedback would be highly appreciated. Thanks!
 
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  • #2
drawar said:

Homework Statement


In a manufacturing process for electrical components, the probability of a finished component being defective is 0.012, independently of all others. Finished components are packed in boxes of 100. A box is acceptable if it contains not more than 1 defective component. Justify the use of a Poisson approximation for the distribution of the number of defective components in a box. 8 boxes are randomly chosen. Given that all of them are acceptable, estimate the conditional probability that they contain exactly 6 defective components altogether.

Homework Equations





The Attempt at a Solution


Let X be the number of defective components in a box containing 100 components, then X~B(100,0.012) => X~P(1.2)
P(X<=1) = 0.663
Let Y be the number of acceptable boxes in 8 randomly chosen ones, thus Y~B(8,0.663)
P(Y=8) = 0.663^8 = 0.0372
P(required) = P(8 acceptable boxes contain 6 defective components) / P(8 acceptable boxes)
I could calculate the probability that 8 randomly chosen boxes contain 6 defective components (using a Poisson distribution with mean=9.6, I guess), but I'm clueless when it comes to '8 acceptable boxes'.
Any feedback would be highly appreciated. Thanks!
You know that the eight boxes are all acceptable, so each has either one or no defective components. So six have to have a defective component. The remaining two have none. What's the probability of that occurring?
 
  • #3
vela said:
You know that the eight boxes are all acceptable, so each has either one or no defective components. So six have to have a defective component. The remaining two have none. What's the probability of that occurring?

So, P(X=1)=0.361 , P(X=0)=0.301
P(8 acceptable boxes contain 6 defective components) = (8C6)((0.361)^6)((0.306)^2))=0.0056
P(required)=0.0056/0.0372=0.151

Am I getting it right?
 
  • #4
Yup, that's it.
 
  • #5
That helps a lot. Thanks
Oh, I have one more question here. If X and Y are 2 independent events, are X and Y', X' and Y' independent also?
(X' and Y' are the complements of X and Y, respectively.)
 

FAQ: Poisson approximation distribution

What is a Poisson approximation distribution?

A Poisson approximation distribution is a probability distribution that is used to approximate the number of events that occur within a specific time period. It assumes that the events occur randomly and independently of each other.

How is a Poisson approximation distribution different from a normal distribution?

A normal distribution is used to model continuous data, while a Poisson approximation distribution is used for discrete data. Additionally, a normal distribution has a bell-shaped curve, while a Poisson approximation distribution has a skewed shape.

What are the assumptions of a Poisson approximation distribution?

The assumptions of a Poisson approximation distribution include that the events occur randomly and independently, the probability of an event occurring is the same for all time periods, and the events cannot occur more than once in a single time period.

How is a Poisson approximation distribution calculated?

The Poisson approximation distribution is calculated using the formula P(x) = (e^-λ)(λ^x)/x!, where λ is the average number of events per time period and x is the number of events that occur in that time period.

What are some real-world applications of the Poisson approximation distribution?

The Poisson approximation distribution is commonly used in fields such as epidemiology, finance, and telecommunications to model the occurrence of rare events, such as disease outbreaks, financial crises, and network failures.

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