At a factory that makes crockery the quality control department has found that 10% of plates have minor faults. These are classed as ‘seconds’. Plates are stored in batches of 12. X~B(12,0.1) where X is the number of seconds in a batch. A random sample of 4 batches is selected. Find the probability that the number of these batches that contain at least 1 second is fewer than 3.
I have tried to answer this question. The probability that a batch has one or more second is: $$P(X\ge1)=1-P(X=0)= 1-({12 \choose 0}\times0.9^{12})=1-0.2824295=0.71757$$ I then supposed that I could then do P(2 seconds)+P(1 second)+P(0 seconds), which I put as: $$a=0.71757$$ $$b=0.28243$$ $$answer= (a^2\times b^2)+(a\times b^3)+(b^4) = 0.0636 (3sf)$$ However, for some reason, I should have used a Binomial distribution again to find the answer, and I should have done: $$answer= ({4\choose2}\times a^2\times b^2)+({4\choose1}\times a\times b^3)+(b^4)=0.317 (3sf)$$ Or in other words I should have used X~B(4,0.28243) and found $P(X\lt3)$. Why did I need a binomial distribution here, why wasn't my method correct. I know that the answer is 0.317 found using the binomial method, and that my 'a' and 'b' values are correct, and I know that everything is correct in my working other than my omission of the 4C2 and 4C1 from my final working of the answer.
Thanks for your help in advance!
Abstractly, because the binomial distribution is the distribution for an experiment counting the number of successes in $n$ Bernoulli trials, which is what you have here. Concretely, because you only counted one possibility of, say, one batch having at least one second and the other three having none, whereas there are $4$ batches and the $4$ scenarios in which a particular one of these $4$ batches has at least one second and the other three don't are mutually exclusive and each have the probability $ab^3$ that you added only once.