## $k$ defects are randomly distributed amongst $n$ integrated-circuit chips produced by a factory (any number of defects may be found on a chip and each defect is independent of the other defects).

Let $p(k, n)$ represent the probability that there is a chip with at least $3$ defects.

For instance $p(3,7) \approx 0.0204081633$.

Find $p(20\,000, 1\,000\,000)$ and give your answer rounded to $10$ decimal places in the form 0.abcdefghij.

### To calculate the probability that there is a chip with at least 3 defects, we can use Poisson distribution, a common distribution used when modeling the number of times an event occurs in an interval of time or space.

The formula of Poisson distribution is:

P(k;λ) = e^-λ (λ^k / k!)

where:

– λ is the average rate of value (the expected number of times the event occurs over the interval), in this case, it is k/n

– k is actual number of successes that result from the experiment

In this case, we are finding the probability that there are three or more defects (k >= 3), or alternatively, that there fewer than three defects (k < 3). Thus we calculate P(0) + P(1) + P(2) where "k" here represents the (less than 3) defects on a chip, and subtract this from 1 (the total probability of all outcomes). Given k=20000 defects amongst n=1000000 chips, λ = k/n = 20000/1000000 = 0.02 P(0;λ) = e^-0.02 * 0.02^0 / 0! = e^-0.02 ≈ 0.980198673 P(1;λ) = e^-0.02 * 0.02^1 / 1! = 0.02e^-0.02 ≈ 0.0196039735 P(2;λ) = e^-0.02 * 0.02^2 / 2! = 0.000196039735 Probability of there being fewer than 3 defects on an individual chip is P(0) + P(1) + P(2) = 0.980198673 + 0.0196039735 + 0.000196039735 = 0.99999868623 In order to transform this into probability of there being at least 3 defects on any chip amongst all chips, we will get, the probability that there is no chip has 3 defects: P'(n) = [P(0) + P(1) + P(2)]^n ≈ (0.99999868623)^1000000 ≈ 0.1352811456 Remember we want the opposite case so we subtract this from 1 giving the final result, p(20000, 1000000) = 1 - P'(n) = 1 - 0.1352811456 = 0.8647188544 So, the probability that there is a chip with at least 3 defects p(20000, 1000000) is 0.8647188544.

##### More Answers:

PrimonacciReflexive Position

Paper-strip Game