Components of a certain type are shipped to a supplier in batches of ten. Suppos

Question

Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50% of all such batches contain no defective components, 30% contain the defective component, and 20% contain two defective components. One component from a batch is randomly selected and tested.

a. What is the probability that the component is not defective?
b. If the sampled component is not defective, what is the probability that it came from a batch with only one defective components?
c. Are the events A={sampled component is not defective} and B={sampled component came from a batch with only one defective component} independent? (please justify your answer with a formula that verifies if two events are independent. State your evidence)

Explanation / Answer

Here's how i solved (a) Let Bk be the event that k defective components are in a batch, and let Tk be the event that k of the two tested components are defective. We seek P(B0 | T0), P(B1 | T0), and P(B2 | T0) First, observing that Bk are mutually exclusive and exhaustive, we may use Bayes' theorem and compute these as: P(Bi | T0)=P(T0 | Bi) P(Bi) /[ P(T0 | B0)P(B0) + P(T0 | B1)P(B1) + P(T0 | B2)P(B2)] look at P(T0 | B0), P(T0 | B1) and P(T0 | B2). P(T0 | B0) = 1, for if our batch contains no defective components, the probability of testing 0 defective components is 1. P(T0 | B1) = (9 choose 2)/(10 choose 2), because we may test 2 out of all but the single defective component, and there are 10 choose 2 ways to test them total. P(T0 | B2) = (8 choose 2)(10 choose 2), because we may test 2 out of all but the two defective components Using these results in the formula for Bayes' theorem, gives me the correct answers for (a). Now for (b), We seek P(B0 | T1), P(B1 | T1), and P(B2 | T1). Using the same method as before, we now have to find the probabilities of P(T1 | B0), P(T1 | B1) and P(T1 | B2) P(T1 | B0) = 0, for if there are no defective components, we cannot test 1 defective one P(T1 | B1) = (9 choose 1)/(10 choose 2), as there are (9 choose 1) ways to select the remaining non-defective component to be tested out of (10 choose 2) ways total P(T1 | B2) = (8 choose 1)/(10 choose 2), as there are (8 choose 1) ways to select the remaining non-defective component. Plugging all this in to bayes' theorem gives me: P(B0 | T1) = 0 P(B1 | T1) = 0.628 P(B2 | T1) = 0.372

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