Yes we have, but the ones we did at school are for example red=0.25 and not red=0.75, if there are 20 cars, what is the probability that 3 are red. But here it's really complex so I'm confused on what to do
I'll stick a sketch solution to Q1 here for now, because it looks a bit annoying. It somewhat requires conditional probability, but it's possible to get away with just a
tree diagram on top of binomial probabilities here.
\[ \text{Seeing as though the lot contains }10\%\text{ defectives,}\\ \text{we anticipate that the probability any randomly selected product is defective}\\ \text{is just }\boxed{p=0.1}. \]
This first bit can be easy to overanalyse, but just break it down logically. 10% of all of them are defective. So why
wouldn't the probability that a randomly selected one be defective, be also equal to 10%?
\[ \text{Taking }n=10\text{ as we choose ten at random,}\\ \text{firstly if none are defective, the batch is accepted without question.}\\ \text{This occurs with probability }\binom{10}{0} (0.9)^{10}. \]
\[ \text{Now, suppose 1 is defective. This occurs with probability }\binom{10}{1} (0.1) (0.9)^9.\\ \text{Conditioned on this, another batch of 10 is taken.}\\ \text{The only way we can have at most 3 in total being defective}\\ \text{is if within the batch of 10, }\textbf{at most 2}\text{ are defective.}\\ \text{This occurs with probability }\left[ \binom{10}{0} (0.9)^{10} + \binom{10}{1} (0.1)(0.9)^9 + \binom{10}{2} (0.1)^2 (0.9)^8 \right]. \]
So that case yields \( \binom{10}{1} (0.1)(0.9)^9 \left[ \binom{10}{0} (0.9)^{10} + \binom{10}{1} (0.1)(0.9)^9 + \binom{10}{2} (0.1)^2(0.9)^8 \right] \)
\[ \text{Similarly, if originally 2 are defective}\\ \text{we have }\binom{10}{2} (0.1)^2(0.9)^8 \left[ \binom{10}{0} (0.9)^{10} + \binom{10}{1} (0.1)(0.9)^9 \right] \]
\[\text{Similarly, if originally 3 are defective}\\ \text{we have }\binom{10}{3} (0.1)^3 (0.9)^7 \binom{10}{0} (0.9)^{10} \]
Arguably a branch-and-conquer method. But it ultimately should add up to give the required answer.