P

Probability

Defect Rate Distribution Calculator

Calculate the exact hypergeometric distribution for defective units in a random sample drawn without replacement from a finite lot.

FINITE-LOT SAMPLE DISTRIBUTION

Enumerate every feasible defect count without replacement

Use a fixed lot composition and a simple random sample. The calculator normalizes stable log-combination weights, reports exact point and tail probabilities, and reconciles the full outcome table to the analytic mean.

P(X = k)-
P(X <= k)-
P(X >= k)-
Expected defects-
Standard deviation-
Feasible support-

LIVE DECISION RECORD

Exact finite-lot outcome distribution

Every feasible sample defect count with exact probability, cumulative probability, upper tail, and distance from the expected count.

Quality technician drawing inspection pieces without replacement from a finite crate while marking possible defect counts
Without replacement, each inspected piece changes the composition left in the lot; the finite-population model captures that dependence.
Exact finite-lot outcome distributionCurrent inputs; comparisons use unrounded values
Every feasible sample defect count with exact probability, cumulative probability, upper tail, and distance from the expected count.
Sample defects xP(X = x)P(X <= x)P(X >= x)Position

CURRENT CALCULATION PROCESS

Formula, current substitution, intermediate values, and reconciliation

P(X=x) = C(D,x) C(N-D,n-x) / C(N,n); E[X] = nD/N

Current symbol, unit, and entered-value register
SymbolMeaning and unitCurrent value
NFinite lot size, units100
DKnown lot defective units8
nSample size without replacement10
kSelected sample defect count2

    Waiting for valid inputs.

    FIVE-STEP USE

    Frame a finite-lot probability question

    1. Confirm the physical lot is finite and sampling removes units from the remaining population.
    2. Enter the lot size and the fixed defect count used by the scenario.
    3. Enter a simple random sample size no larger than the lot or 5,000-unit reporting limit.
    4. Choose an exact defect count and decide whether the point, lower-tail, or upper-tail question matches the decision.
    5. Export the complete outcome table and retain the randomization procedure with the lot record.

    FIVE FINITE-LOT FUNDAMENTALS

    Why the distribution changes after every draw

    Without replacement
    An inspected unit is not returned before the next selection.
    Fixed composition
    The scenario conditions on exactly D defective units in N total units.
    Feasible support
    Counts run only from `max(0,n-(N-D))` through `min(n,D)`.
    Point probability
    The mass assigned to exactly one sample defect count.
    Finite correction
    Variance shrinks as the sample consumes a larger share of the lot.

    FORMULA AND DEFAULT SUBSTITUTION

    Trace a 10-unit sample from a 100-unit lot

    P(X=2) = C(8,2) C(92,8) / C(100,10) = 0.15056142397

    For `N=100`, `D=8`, and `n=10`, the expected count is `10 x 8/100 = 0.8`. Summing masses through 2 gives 96.8612%; summing from 2 upward gives 18.1950%. The overlap at exactly 2 belongs to both inclusive tails.

    THREE DEEPER MODULES

    Read the exact table in context

    Sampling fraction

    When the sample is tiny relative to the lot, a binomial approximation may be close. As the fraction grows, without-replacement dependence materially reduces variance.

    Inclusive-tail overlap

    `P(X<=k) + P(X>=k)` equals `1 + P(X=k)`, not one. This is a useful reconciliation and prevents treating overlapping tails as complements.

    Lot-composition scenario

    If D is not known, rerun plausible compositions. The distribution then shows conditional detection power rather than estimating the unknown lot defect count.

    TWO SAMPLING CASES

    Partial and complete inspection

    Supplier containment sample

    From a 100-unit lot believed to contain 8 defectives, a 10-unit draw has a nontrivial chance of seeing none or one. The table makes the miss risk visible before a containment rule is approved.

    Inspect the whole lot

    With `n=N`, the sample defect count must equal D. Variance becomes zero and the outcome table collapses to one row with probability 100%.

    FINITE-LOT GLOSSARY

    Six terms for the sampling record

    Lot
    The bounded set from which units are selected.
    Simple random sample
    Selection giving eligible samples equal probability.
    Hypergeometric variable
    Defect count in a fixed-size sample without replacement.
    Combination
    The number of unordered ways to select a subset.
    Cumulative probability
    Total mass at or below a selected count.
    Upper tail
    Total mass at or above a selected count.

    LIMITS AND EVIDENCE

    Random selection is part of the model

    • The lot composition is fixed and supplied; it is not inferred from the sample.
    • Sampling must be without replacement and reasonably random across the lot.
    • Inspection misclassification, clustering, stratification, and process drift are excluded.
    • Samples above 5,000 are blocked because the page reports the complete exact support.
    • The result is a conditional probability calculation, not a regulatory acceptance plan.

    Retain: lot identity, N and D basis, randomization method, sample identifiers, inspection disposition, selected tail rule, and exported exact table.

    RELIABLE SOURCES

    Primary finite-sampling references

    DEFECT DISTRIBUTION FAQ

    Questions about support, tails, and replacement

    Why is this not a binomial distribution?

    The finite lot is sampled without replacement, so each draw changes what remains. The hypergeometric model includes that dependence and finite-population correction.

    What does known lot defects mean?

    It is the fixed defective count assumed to exist in the entire lot for this scenario. If it is unknown, the page is a what-if model, not an estimate.

    Why can some sample counts be impossible?

    A sample cannot contain more defects than the lot holds or more conforming units than the lot holds. The feasible support enforces both constraints.

    Which tail should I use?

    `P(X>=k)` answers at least k defects; `P(X<=k)` answers at most k. The point probability answers exactly k and should not be confused with either tail.

    Why limit the sample to 5,000 units?

    The page prints every feasible exact outcome for review. The limit protects browser responsiveness and PDF readability while allowing large finite lots.

    Does a small tail probability prove the lot composition is different?

    Not by itself. The result is conditional on simple random sampling and the assumed fixed lot composition; inspection errors and selection bias can dominate.

    IMPORTANT SAMPLING NOTE

    An exact formula cannot repair a biased draw

    The probabilities are exact for the stated finite-lot model. A convenience sample from accessible cartons can still make the operational conclusion unreliable.