Consider a machine that produces items in sequence. Under normal operating conditions, the items are independent with a probability of 0.01 of being defective. However, it is possible for the machine to develop a “memory” in the following sense: After each defective item, and independent of anything that happened earlier, the probability that the next item is defective is\(\frac{{\bf{2}}}{{\bf{5}}}\). After each non-defective item, and independent of anything that happened earlier, the probability that the next item is defective is\(\frac{{\bf{1}}}{{{\bf{165}}}}\).
Please assume that the machine is operating normally for the whole time we observe or has a memory for the whole time we observe. LetBbe the event that the machine is operating normally, and assume that\({\bf{Pr}}\left( {\bf{B}} \right){\bf{ = }}\frac{{\bf{2}}}{{\bf{3}}}\). Let\({{\bf{D}}_{\bf{i}}}\)be the event that theith item inspected is defective. Assume that\({{\bf{D}}_{\bf{1}}}\)is independent ofB.
a. Prove that\({\bf{Pr}}\left( {{{\bf{D}}_{\bf{i}}}} \right){\bf{ = 0}}{\bf{.01}}\) for alli. Hint:Use induction.
b. Assume that we observe the first six items and the event that occurs is\({\bf{E = D}}_{\bf{1}}^{\bf{c}} \cap {\bf{D}}_{\bf{2}}^{\bf{c}} \cap {{\bf{D}}_3} \cap {{\bf{D}}_4} \cap {\bf{D}}_{\bf{5}}^{\bf{c}} \cap {\bf{D}}_{\bf{6}}^{\bf{c}}\). The third and fourth items are defective, but the other four are not. Compute\({\bf{Pr}}\left( {{\bf{B}}\left| {\bf{E}} \right.} \right)\).
