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BINOM.INV in WPS Spreadsheet: Find the First Binomial CDF Crossing

Statistical Intermediate WPS Spreadsheet

Find the smallest success count reaching a cumulative probability threshold with explicit binomial checks and documented scope.

Verification status: See the evidence and test scope described in this guide. How formula examples are checked

Quick answer

=BINOM.INV(10,0.2,0.8) has independently expected output 3. It returns the smallest integer success count k for which the binomial cumulative probability P(X ≤ k) reaches or exceeds alpha. It does not return a probability or interpolate between counts.

Syntax and arguments

BINOM.INV(trials,probability_s,alpha)

All arguments are required:

  • trials: number of Bernoulli trials. Use a nonnegative integer; the source says fractional trial counts are truncated.
  • probability_s: success probability on each trial, between 0 and 1.
  • alpha: cumulative threshold, between 0 and 1 in the documented error checks. The examples use interior values.

The binomial model assumes a fixed trial count, independent trials, two outcomes per trial and the same success probability throughout. Define what counts as a success before calculating.

Worked worksheet: ten independent checks

Suppose each of ten checks has a model probability of 0.2 of finding an issue. Put 10 in B2, 0.2 in B3 and 0.8 in B4; label these Trials, Issue probability and Cumulative threshold. In D2 enter =BINOM.INV(B2,B3,B4).

Expected output: 3 issues. Independent cumulative probabilities are P(X ≤ 2) = 0.6777995264 and P(X ≤ 3) = 0.8791261184. The threshold 0.8 falls between them, so 3 is the first qualifying count. This does not mean the probability of exactly three issues is 80%.

Additional threshold checks

Formula Expected count
=BINOM.INV(10,0.2,0.5) 2
=BINOM.INV(10,0.2,0.9) 4
=BINOM.INV(10,0.2,0.95) 4

Several thresholds can return the same count because the cumulative distribution advances in discrete steps. The last two both cross at 4, whose cumulative probability is 0.9672065024.

Advanced example: equality at a cumulative step

For four trials with success probability 0.5, the cumulative probability through one success is (1+4)/16 = 0.3125.

=BINOM.INV(4,0.5,0.3125)
=BINOM.INV(4,0.5,0.32)

Expected results: 1 and 2, respectively. Equality qualifies; moving just above a step can increase the integer output. This exact dyadic-probability check avoids relying on a rounded decimal approximation to a threshold.

Independent verification sums comb(n,j)*p^j*(1-p)^(n-j) from j = 0 through each candidate count. The preceding count must be below alpha, and the chosen count must reach it.

Troubleshooting and limitations

The source lists #VALUE! for nonnumeric arguments and #NUM! for negative trials, success probability outside [0,1], or alpha outside [0,1]. It does not establish all degenerate endpoint combinations or current numerical behavior for very large n; those remain untested.

If a count surprises you, check whether 20 was entered instead of 20%, or whether success was defined as a pass in one place and a failure elsewhere. Replacing varying probabilities by their mean does not generally make dependent or unequal-probability trials binomial.

The edition banner is empty, so no particular Windows edition is claimed here. A related CRITBINOM lesson exists, but the reviewed evidence does not explicitly establish a rename or universal alias relationship.

Availability and verification

The official WPS BINOM.INV lesson was fetched afresh on October 19, 2026. Its edition banner is empty. The Chinese WPS Academy worksheet demonstration supports this function identity, but does not establish a particular edition. The lesson dates from 2019; it does not establish introduction or retirement dates, exact build coverage, or current Mac, Linux, web and mobile behavior. Examples are independently calculated expectations, not observed WPS outputs. Native execution and genuine current-application screenshots remain unverified.