Expected items until two defects in a row

Question

A machine produces items that are independently defective with probability $\tfrac15$ . An inspector watches the stream and stops the moment two defective items appear back to back.

What is the expected number of items inspected until the inspector stops?

Accepted Answer

Two states: A = last item good (or start), B = last item defective. With defect probability p = 	frac15, eA = 1 + 	frac15 eB + 	frac45 eA, \qquad eB = 1 + 	frac15(0) + 	frac45 eA. The first gives 	frac15 eA = 1 + 	frac15 eB, so eA = 5 + eB. Substitute: eB = 1 + 	frac45(5 + eB) = 5 + 	frac45 eB, so 	frac15 eB = 5, eB = 25, and eA = 30. Fresh start: eA = 30. Takeaway. Two defects in a row average \frac{1+p}{p^2} = \frac{6/5}{1/25} = 30 items when p = 	frac15.

Expected items until two defects in a row

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