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Library / Foundations: Maths / FMA-31

Foundations: Maths

Binomial probability

Chance of exactly k successes in n independent trials.

Ref
FMA-31
Rev
v1
Field
Foundations: Maths
Result
P(X=k) [%]

Use n independent trials with the same success probability p and two outcomes. Mean = np and variance = np(1 − p).

Count the ways to arrange k successes among n trials, and multiply by the chance of each arrangement.

Calculator

Solve for
%
220
010
0.050.5
  • Mean np1.5
  • Standard deviation √(np(1−p))1.025

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Probability (%) current value 30.87 %
kNumber of successesKey driver
0.24336
pSuccess probability
2.1434.6
nNumber of trials
2.7832.4
  • The most likely number of successes is close to np.
  • Use it for repeated firings, test trials or component checks where each outcome is pass or fail.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (%)Local stepChange in result
k Number of successes0 to 10 0.24336.0135.77+10%−7.11 pp
p Success probability0.05 to 0.5 2.14334.5532.41+undefined%—
n Number of trials2 to 20 2.78532.4429.65+10%+1.09 pp

Sweep of Success probability

Success probabilityProbability (%)
0.052.143
0.10638.059
0.162515.51
0.218822.82
0.27528.82
0.331232.82
0.387534.5
0.443733.89
0.531.25

Sweep one variable

Probability (%)Hover, touch, or use the arrow keys on the chart
0102030 0.10.20.30.40.5 Success probability now 30.87

Variables and how each one contributes

P(X=k)

Probability

Result · %

What the equation gives you. Everything else feeds it.

n

Number of trials

· typical 2 to 20

How many trials. More trials spread probability across more outcomes.

k

Number of successes

· typical 0 to 10

The count you ask about. Probability peaks near np.

p

Success probability

· typical 0.05 to 0.5

Chance of success in each trial. Strong: probability peaks when p = k/n.

Worked example with the values above

  1. Ways to choose: C(5, 2) = 10.
  2. Chance of one arrangement: 0.32 × 0.73 = 0.03087.
  3. P = 30.87 %.

Assumptions and limits

  • Independent trials with a constant p.
  • Exactly two outcomes per trial.

Common mistakes

  • Forgetting the C(n, k) factor.
  • Using it when trials depend on each other.

Statistics and uncertainty

The standard deviation √(np(1−p)) tells how far results typically stray from np; observed frequencies converge on p as n grows, with error about √(p(1−p)/n).

Where it is used

  • Probability of k hits in n shots.
  • Number of failures in a batch of components.

Maths you need

  • Combinations: n choose k
  • Independent events: multiply probabilities

Related formulas

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