Question

Consider a software program that is artificially seeded with 100 faults. While testing this program, 159 faults are detected, out of which 75 faults are from those artificially seeded faults. Assuming that both real and seeded faults are of same nature and have same distribution, the estimated number of undetected real faults is _____.

Official answer and solution
Official answer
28
From
Official GATE 2015 answer key

Solution

Apply Mills' fault-seeding model (capture-recapture estimation) to compute the total and undetected real faults.

Under Mills' fault-seeding model, artificially seeded faults are assumed to have the same probability of detection as inherent real faults.

Let:

  • S=100S = 100: number of artificially seeded faults
  • s=75s = 75: number of seeded faults detected
  • Total detected faults =159= 159
  • rr: number of real faults detected =159−75=84= 159 - 75 = 84
  • NN: total estimated number of real faults in the software

Equating the detection ratios:

sS=rN  ⟹  75100=84N\frac{s}{S} = \frac{r}{N} \implies \frac{75}{100} = \frac{84}{N}
0.75⋅N=84  ⟹  N=840.75=84×43=28×4=1120.75 \cdot N = 84 \implies N = \frac{84}{0.75} = \frac{84 \times 4}{3} = 28 \times 4 = 112

Total estimated real faults originally present is 112.

The question specifically asks for the estimated number of undetected real faults:

Undetected real faults=N−r=112−84=28\text{Undetected real faults} = N - r = 112 - 84 = 28
Final answer

28

Common trap

Reporting total real faults (112) instead of the undetected real faults (28 = 112 - 84).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2015 Computer Science and Information Technology (CS) Session 3
Question
21 of 65, printed as CS Q.11
Section
Computer Science and Information Technology

This question is from a part of the syllabus that is no longer in GATE CS. It is kept because it was in the paper.

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