Consider a machine that produces items in sequence. Under normal operating condi

Question

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 nondefective 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. Prove that Pr(D_i) = 0.01 for all Assume that we observe the first six items and the event that occurs is E = D_1^c n D_2^c n D_3 n D_4 n D_5^c n D_6^c. That is, the third and fourth items are defective, but the other four are not. Compute Pr(B|E).

Explanation / Answer

Solution

Back-up Theory

If an event D is dependent on the outcome of another event C, say the outcomes of C are C1, C2,C3, ….., Ck, then

P(D) = P(D/C1)xP(C1) + P(D/C2)xP(C2) + P(D/C3)xP(C3) + ….. + P(D/Ck)xP(Ck)

Now, to work out the solution, Part (a)

P(D2) =

P(D2/D1)xP(D1) + P(D2/D1’)xP(D1’) where D1’ is the complement of D1, i.e., not defective

= 0.4x0.01 + (0.99/165) = 0.004 + 0.006 = 0.01.

So, P(D2) = 0.01…… (1)

Now, to go for induction, let P(Dk) = 0.01.

Then, P(Dk+1) = P(Dk+1/Dk)xP(Dk) + P(Dk+1/Dk’)xP(Dk’)

= 0.4x0.01 + (0.99/165) = 0.004 + 0.006 = 0.01.

So, if P(Dk) = 0.01, then P(Dk+1) = 0.01 …. (2)

(1) and (2) by Principle of Induction => P(Di) = 0.01 for all integer values of i. PROVED

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