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?
Best Answer
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).