Consider a machine that produces items in sequence. Under normal operating conditions, the items are independent with probability 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 2/5. After each non-defective item, and independent of anything that happened earlier, the probability that the next item is defective is 1/165. Assume that the machine is either operating normally for the whole time we observe or has a memory for the whole time that we observe. Let B be the event that the machine is operating normally, and assume that Pr(B) = 2/3. Let $D_i$ be the event that the $i^{th}$ item inspected is defective. Assume that $D_1$ is independent of B. Assume that we observe the first six items and the event that occurs is E = $D^c_1 \cap D_2^c \cap D_3 \cap D_4 \cap D_5^c \cap D^c_6 $. That is, the third and fourth items are defective, but the other four are not. Compute Pr(B|D)
I would like to know what the D represents in Pr(B|D). Does it mean that knowing the next item is defective, what's the probability that the machine operates normally? And besides what's the final answer and procedure? (it's no problem that the first question mentioned $Pr(D_i)$ = 0.01 for all i using induction.)