# Poisson approximation distribution

## 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.

## 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!

vela
Staff Emeritus
Homework Helper

## 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.

## 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?

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?

vela
Staff Emeritus