KCC's Quizzes AQQ244 about Wafer Processing Yield puzzle

I assume that ATE tester has only two states:  Good,  or Bad. (a third one, inconclusive, is not considered).

Assume x is the probability that the IC is really good, so 1-x  is the probability that the IC is really bad  ( 0% = other that good or bad).

1. When the IC is really good, ATE reports it good in 96% of the cases, (so bad, in 4%, since ATE has only two possible conclusions).

    Good and bad (in this case)  = x,  0.96 x = reported as good,   0.04 x = reported as bad (while good).

2. When the IC is effectively bad,  ATE reports is as such 98%, so reports is as good in 2% of these cases:

    Bad IC, (1-x),  reported as such:  0.98 * (1-x)  while bad but reported good 0.02*(1-x)

0. Assumed yield is the good reported good  +  the bad reported good  = 0.96x  + 0.02(1-x)  = 0.95   (I assume this how the "yield" is defined; another possible definition would be x = 0.95)

    which leads to  x = 93 / 94

Q1: ATE is wrong by reporting bad while the IC is good  +  reporting good while the IC is bad : 0.04x + 0.2*(1.x)  =  3.9798%   (almost 4%)

Q2: ATE reporting good a bad IC:  0.2(1-x)  = 0.0212%  ( or one IC each 4700)

An easier way to “visualize” the problem (without conditional probability, but by simple algebra):

Define the 4 variables p, q, r and s:

The 4 linear equations are:

p + q + r + s  = 1     // (no other case, so total is at 100%)

p = 0.96 ( p+r )  //  since p+r = total of IC is good, p = detected good at 96%

s = 0.98 ( q+s )   //  since q+s  = total of IC is bad, s = detected as bad at 98%

Missing one equation to solve the four unknowns. Given by the “yield” which, I don’t know, can be p+r  or p+q. One of these sums must be equals to 0.95.

With p+r = 0.95, we obtain the system of equations:

Gives: r=19/500, s=49/1000, q=1/1000, p= 114/125

While if it is p+q  which is the yield

Gives: r=93/2350, s=49/4700, q=1/4700 and p=116/1175

(solutions from WolframAlpha, if there is an error, it is probably from my transcription)

The answer of Q1 is  q+r  while the answer of Q2  is just q.

Actual yield (good chips) = 0.95
Bad chips = 0.05
Good chips that ATE confirms good = 0.96 * 0.95 = 0.912
Good chips that ATE rejects = 0.04 * 0.95 = 0.038
Bad chips that ATE rejects = 0.98 * 0.05 = 0.049
Bad chips that ATE misses (tests good) = 0.02 * 0.05 = 0.001

1) Probability that ATE tester is wrong = 0.038 + 0.001 = 0.039 [or 3.9%]
2) Probability that an IC (tested good) is in fact bad = 0.001 [or 0.1%]

Big applause Martin! All answers are correct!

Thanks vanderghast , thus from your table, the probability that the ATE is wrong is 4% and that a BAD device is declared good is 0.1% ? I have well the 4% answer but not the second one. Can you recheck?

Indeed,  with the last equation corrected, it should be  (1 = 1%):  

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We can assume roughly 5% of defective chips (yield), as most defects are found.

1. Of the 95% roughly good ICs, it directly finds 96%= 0.95*0.96=91.2%. Of the 5% defective is directly finds 98%. This makes up for 5%*98%=4.9%. Thus, is is wrong in 100%-4,9%-91.2%=3.9% of the results. 

2. The OK test rate of 96% plays a minor role in IC test, because chips that are not OK, e.g. due to a contact failure, typically become re-tested. 

With the ATE only recognizing 98% of defect ICs, this means, that 2% of the roughly 5% defective ICs are not discovered and therefore 0,02*0,05=0.1% of the ICs tested "good" still have an undiscovered fault. 

We can assume roughly 5% of defective chips (yield), as most defects are found.

1. Of the 95% roughly good ICs, it directly finds 96%= 0.95*0.96=91.2%. Of the 5% defective is directly finds 98%. This makes up for 5%*98%=4.9%. Thus, is is wrong in 100%-4,9%-91.2%=3.9% of the results. 

2. The OK test rate of 96% plays a minor role in IC test, because chips that are not OK, e.g. due to a contact failure, typically become re-tested. 

With the ATE only recognizing 98% of defect ICs, this means, that 2% of the roughly 5% defective ICs are not discovered and therefore 0,02*0,05=0.1% of the ICs tested "good" still have an undiscovered fault.