Probability of finding P(X=k)? A factory produces 10 glass containers daily. It may be assumed that there is a constant probability p=0.1 of producing a defective container. Before these containers are stored they are inspected and the defectives ones are set aside. Suppose that there is a constant probability r=0.2 that a defective container is misclassified. Let X equal the number of containers classified as a defective at the end of a production day. (Suppose that all containers which are manufactured on a particular day are also inspected on that day.)
(1)P(X=k) = ?
 A: The wording of the problem is unfortunately ambiguous. But we lean to the interpretation that if a container is defective, there is a probability  $0.2$ that the containder is classified as non-defective.  Nothing is said about classification errors on non-defectives. So we will assume that a non-defective is always classified as non-defective.
Thus the probability that a container is set aside is $(0.1)(0.8)=0.08$. Call this number $p$.
Then the number of containers set aside has binomial distribution. The probability that this number is $k$ is $\dbinom{10}{k}p^k(1-p)^{10-k}$.
Remark: The problem could be interpreted as meaning that with probability $0.2$, a container is misclassified. (But the problem does not say that.) In that case, a container can be set aside for two disjoint reasons: (i) It is defective, and correctly set aside or (ii) it is OK, but wrongly set aside. Then the probability of being set aside is $(0.1)(0.8)+(0.9)(0.2)=0.26$. The rest of the calculation would be as above, with $p=0.26$.
