Probability Defective Items Binomial Distribution

Question

Suppose that a large lot with 10000 manufactured items has 30 percent defective items and 70 percent nondefective. You choose a subset of 10 items to test. (a) What is the probability that at most 1 of the 10 test items is defective? (b) Approximate the previous answer using the binomial distribution.

I am getting for (a) that $P(\text{at most 1 def item}) = 0.7^{10} + {10\choose 1} \cdot 0.3^1 \cdot 0.7^9$

I do not understand what is meant by (b), since the answer for (a) already uses binomial distribution?

Answer 1

The $10$ items are not chosen independently since they are chosen without replacement. If it were with replacement, with a tiny chance that the same item might be chosen more than once, then the binomial distribution would be exact rather than a very close approximation. As it is, part (a) must use a hypergeometric distribution. The answer you've written for part (a) is in fact a correct answer to part (b).

Answer 2

The binomial distribution is only an approximation

More explicitly

$P(\text{at most 1 defective }) = P(\text{ no defective item}) + P(\text{1 defective item}) \\= \frac{7000}{10000}\frac{6999}{9999}\ldots \frac{6991}{9991} + 10 \frac{3000}{10000}\frac{7000}{9999}\frac{6999}{9998} \ldots \frac{6992}{9991} $