Question

supposed that in a large population of manufactured items the proportion of defective items is 0.008 . what is the probability that in a random sample of 2000 items there will be 4 or more defective items
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Answer 1

Given:

No. of items in the sample = 2000

The proportion of defective items in the population = 0.008 = $\frac{0.8}{100}$ = 0.8%

To find:

The probability that there will be 4 or more defective items in the sample.

Solution:

Here, we have a large population of items and we select 2000 random items out of which 0.8% are defective. We apply binomial distribution where we apply a random experiment of choosing 4 or more defective items. Let $r$ be the no. of successes in $n$ no. of trials.

Then, the probability of $r$ successes in $n$ trials in a specific order is given by the formula: $p^{r}q^{n-r}$

where $p$ is the probability of a success and $q=1-p$ is the probability of a failure in a trial. If $r$ successes in $n$ trials occur in $nC_{r}$ ways and the probability for each of the ways is $p^{r}q^{n-r}$, then the probability of $r$ successes in $n$ trials is given by the formula:

$P(r)=nC_{r}p^{r}q^{n-r}$

where $p+q=1$ and $r=0,1,2...n$

Here, success $(p)$ is getting a defective item and failure $(q)$ is not getting a defective item.

Thus, the no. of successes is 4 or more, the probability of the defective item in the population is 0.8% and the no. of sample items is 2000. Hence,

$p=\frac{0.8}{100}, n=2000, r=4$

$q=1-p$

$q=1-\frac{0.8}{100}= \frac{99.2}{100}$

Now, probability of getting 4 or more defective items in the sample, $P(r)=nC_{r}p^{r}q^{n-r}$

$P(4)=2000C_{4}(\frac{0.8}{100})^{4}(\frac{99.2}{100})^{2000-4 }$

$P(4)=\frac{2000!}{4!(2000-4)!}*\frac{0.8^{4}}{100^{4}} *\frac{99.2^{1996}}{100^{1996}}$

This is the probability of finding 4 or more defective items in the sample.

The probability of finding 4 or more defective items in a random sample of 2000 manufactured items is $P(4)=\frac{2000!}{4!(2000-4)!}*\frac{0.8^{4}}{100^{4}} *\frac{99.2^{1996}}{100^{1996}}$