There are $5\%$ defective items in bulk. What is the probability that a sample of $10$ items includes not more than one defective item?

Question

In my textbook as well as a test, I came across this probability question:

There are $5\%$ defective items in a large bulk of items. What is the probability that a sample of $10$ items will include not more than one defective item?

And I thought of applying binomial distribution as below: $$P(X=r)= {n \choose r}.[P(X)]^r.[P(\bar X)]^{n-r}$$ where $X$ is the event that a defective item is drawn, $r$ is the number of times the event is repeated, $n$ is the number of items drawn.

But then I remembered that for this to work, the experiment needs to be repeated a finite number of times and the probabilities of our success and failure cannot change throughout the process, which is what will happen here each time I draw an item without replacement.

But the solution actually turns out to be exactly this, in which they have considered that the probability remains constant due to the pool of items being large(I didn't get any marks for my ragged attempt to get an exact answer, which I left in uncalculated form because it was ridiculously long).

This doesn't sit well with me as they don't tell us to approximate in the question.

Is there another way where we can find the actual exact answer to this question?

Answer 1

So in your comment you are saying that the lot is $100$, the number of defectives is $5$, and you seek $P(X\leq1)$

As the numbers keep on changing, the exact formula will be

$P(X\leq 1) = \dfrac{\binom 5 0\binom{95}{10}+\binom51\binom{95}9}{\binom{100}{10}} = \dfrac{43877}{47530}$

This is called a hypergeometric distribution.

Answer 2

One way of doing this problem is to use Poisson distribution as the probability of success (defective) is small.

$p = 5 \% = 0.05$.

$n = 10$ (sample size)

So, the parameter $\lambda$ for Poisson distribution is

$\lambda = n p = 10 \times {5 \over 100} = 0.5$

Required probability = $P(X \leq 1) = P(X = 0) + P(X = 1)$.

Using the Poisson distribution,

$P(X = 0) = e^{-0.5} = 0.6065$.

$P(X = 1) = (0.5) \times e^{-0.5} = 0.3033$

Required probability = $P(X = 0) + P(X = 1) = 0.9098$.