How many inspections are needed when parts are randomly arranged.

Inspection count, random part arrangement, and probability of no inspection

This post intends to calculate of how many inspections are needed to meet a false pass percentage when the parts are randomly arranged. The probability that the part will be the correct side toward the camera, not blocked by double stacking, or other unknown reason are considered. It is assumed that the part arrangement is randomized before each inspection.

Calculating the minimum number of inspections is based on the average probability that the part will be arranged in a way suitable for inspection. Similar to the question, “how many times must a coin be flipped to get heads?”

The probability (p) of getting heads is 0.5 (50%).

The negative result (tails) probability (q) is 1 – p = 1 – 0.5 = 0.5 (50%).

Average flips (inspections) needed = 1 / p = 1 / 0.5 = 2

The minimum part inspections must considered probability of wrong side up, part double stack, and unknown random error.

Like the coin flip, there is a 50% (0.5) chance of wrong side.

The double stacking must be mechanically limited to 10% (0.1),

and the unknown error is guessed to be 5% (0.05) [it's unknown].

These errors are independent, the probabilities are additive.

The probability the part can't be inspected is 65% (q = 0.50 + 0.10 + 0.05 = 0.65).

The probability that the part can be inspected is 1.0 – 0.65 = 0.35 = p = 35%

Only 35% of the randomly arranged parts will be able to be inspected.

The minimum number of inspections needed is 1/p = 1/0.35 = 2.857 or 3 inspections.

Now considered the likelihood that the part fails to be inspected 2 times in a row. The sequential probability is q^ˣ, where x is the number of repeated inspections.

If there is a 65% chance the part is not inspected (q = 0.65). Then randomizing the part arrangement and repeating the test gives a 42.25% chance the part isn't inspected (q^² = 0.65^² = 0.4225). Adding more inspections continues to reduce the chance that the part is not inspected. After the 7^th inspection, there is less than a 5% chance a part has not been inspected. By the 11^th inspection, less than 1% have not been inspected.

How many inspections are needed depends on the maximum allowable False Pass Rate and the true Process Defect Rate. If 1 part in 10,000 is allowed to be bad the False Pass Rate is 0.01% (0.0001). This is 0.01% of all parts. If the Process Defect Rate is 2%, for every 10,000 parts 200 are bad. The “No Inspection” parts are assumed to have the same evenly distributed 2% defect rate. The maximum allowable “No Inspect” rate = False Pass Rate / Process Defect Rate = 0.0001 / 0.02 = 0.005 (0.5%). Based on the chart below, 13 inspections are needed.

To calculate the number of inspections required for randomly arranged parts:

Inspection Count =

log( False Pass Rate / Process Defect Rate) / log( Probability No Inspection)

Where:

False Pass Rate = percentage of maximum allowable bad parts getting through the process.

Process Defect Rate = percentage of bad parts made by the process.

Probability No Inspection = the sum of all error probabilities that would prevent the part from being inspected. [In this example, Wrong side up + Double stack + unknown error.]

For the question “Is it cheaper to add more inspections, or to add material handling?”

This calculation is useful when estimating the expense of random arrangement inspections versus material handling positive part presentation.

 20180228 Lowell Cady

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