Probability and distributions

Why this matters

“95% mission success” sounds good, but if a unit flies 20 missions a week, some week is very likely to include a failure. Probability turns risk figures into terms that match real use, and the distribution of data answers questions such as “what is the chance this battery flies less than 20 minutes?”, which is the basis of testing in module 3.

Basic rules of probability

A probability is a number from 0 to 1 that says how likely an event is.

• Complement: $P(\text{not }A)=1-P(A)$
• Independent events: one outcome does not change the chance of the other, so probabilities multiply: $P(A\cap B)=P(A)P(B)$
• Conditional probability $P(A\mid B)$ is the chance of $A$ given that $B$ has happened

The chance that both flights succeed is $0.95\times 0.95=0.9025$, and the four leaves always add up to 1. Multiplying like this only works when flights really are independent. If failures share a cause, such as a windy day or a bug in one software version, flights are not independent.

The binomial distribution

When a trial is repeated $n$ times, each ending in success or failure with a fixed chance, and each trial is independent, the number of failures $X$ follows a binomial distribution.

from scipy import stats

n, p_fail = 20, 0.05
print(f"P(at least one failure in {n} flights) = {1 - 0.95 ** n:.4f}")
failures = stats.binom(n, p_fail)
for k in range(4):
    print(f"P(X = {k}) = {failures.pmf(k):.4f}")
print(f"P(X >= 2) = {failures.sf(1):.4f}   expected failures = {failures.mean():.1f}")

P(at least one failure in 20 flights) = 0.6415
P(X = 0) = 0.3585
P(X = 1) = 0.3774
P(X = 2) = 0.1887
P(X = 3) = 0.0596
P(X >= 2) = 0.2642   expected failures = 1.0

Even though each flight succeeds 95% of the time, 20 flights have a 64% chance of at least one failure, and one failure is expected on average. That helps plan spares and emergency procedures.

p_fault = 0.02
p_alarm_if_fault = 0.90
p_alarm_if_ok = 0.05
p_alarm = p_alarm_if_fault * p_fault + p_alarm_if_ok * (1 - p_fault)
p_fault_if_alarm = p_alarm_if_fault * p_fault / p_alarm
print(f"P(alarm) = {p_alarm:.4f}   P(fault | alarm) = {p_fault_if_alarm:.3f}")

P(alarm) = 0.0670   P(fault | alarm) = 0.269

The normal distribution

Data produced by many small effects added together, such as endurance influenced by wind, temperature and battery condition, often looks like the normal distribution, a symmetric bell curve defined by its mean $\mu$ and standard deviation $\sigma$.

A z-score $z=(x-\mu )/\sigma$ says how many $\sigma$ a value lies from the mean. For a normal distribution about 68.27% of data lies within $\pm 1\sigma$, 95.45% within $\pm 2\sigma$ and 99.73% within $\pm 3\sigma$: the 68–95–99.7 rule.

for k in (1, 2, 3):
    print(f"within ±{k} sigma: {stats.norm.cdf(k) - stats.norm.cdf(-k):.4%}")

mu, sigma = 23.0, 1.5
print(f"z of 20 min = {(20 - mu) / sigma:.1f}   P(endurance < 20 min) = {stats.norm.cdf(20, mu, sigma):.4f}")
print(f"99% of flights last at least {stats.norm.ppf(0.01, mu, sigma):.2f} min")

within ±1 sigma: 68.2689%
within ±2 sigma: 95.4500%
within ±3 sigma: 99.7300%
z of 20 min = -2.0   P(endurance < 20 min) = 0.0228
99% of flights last at least 19.51 min

If a battery model’s endurance is normal with mean 23 minutes and SD 1.5 minutes, the 20-minute requirement lies 2 SD below the mean, so a flight falls short about 2.3% of the time. norm.cdf gives the area to the left of a value; norm.ppf does the reverse, finding the value for a given area.

Monte Carlo simulation and the central limit theorem

When formulas get hard, we simulate the situation with many random numbers and count the outcomes: Monte Carlo simulation. Always set a seed for the random number generator so results repeat.

import numpy as np

rng = np.random.default_rng(116)
flights = rng.normal(mu, sigma, size=100_000)
print(f"simulated P(< 20 min) = {np.mean(flights < 20):.4f}   exact = {stats.norm.cdf(20, mu, sigma):.4f}")

pairs = rng.normal(mu, sigma, size=(100_000, 2))
both_ok = np.mean((pairs >= 20).all(axis=1))
print(f"two batteries both reach 20 min: simulated {both_ok:.4f}   exact {(1 - stats.norm.cdf(20, mu, sigma)) ** 2:.4f}")

simulated P(< 20 min) = 0.0234   exact = 0.0228
two batteries both reach 20 min: simulated 0.9550   exact 0.9550

The simulation is close to the formula and gets closer with more runs. The method works for far more complex problems, such as multi-stage missions where each stage has its own distribution.

The central limit theorem says that when we average enough independent samples, the mean is close to normally distributed even if the original data is not, and its standard deviation shrinks to $\sigma /\sqrt{n}$.

import pandas as pd

lat = pd.read_csv("latency.csv")["latency_ms"].to_numpy()
means = rng.choice(lat, size=(20_000, 25)).mean(axis=1)
print(f"data: sd {lat.std(ddof=1):.2f} ms, skew {stats.skew(lat):.2f}")
print(f"means of 25 frames: sd {means.std(ddof=1):.2f} ms (sigma/sqrt(25) = {lat.std(ddof=1) / 5:.2f}), skew {stats.skew(means):.2f}")

data: sd 11.41 ms, skew 0.78
means of 25 frames: sd 2.30 ms (sigma/sqrt(25) = 2.28), skew 0.18

The processing times are right-skewed (clearly positive skew), but means of 25 frames are much less skewed and about five times less spread, as $\sigma /\sqrt{25}$ predicts. This is why module 3 can build confidence intervals from the distribution of the mean.

Module lab

Common mistakes

Summary

• Probabilities of independent events multiply; conditional probabilities need Bayes’ theorem
• The binomial distribution counts successes or failures in repeated independent trials
• The normal distribution is set by $\mu$ and $\sigma$; z-scores and the 68–95–99.7 rule apply when data is close to normal
• Monte Carlo simulation needs a seed, and sample means approach normality under the central limit theorem

Check your understanding

1. Three drones fly at once, each with an independent 2% chance of a fault. What is the chance that none has a fault?
2. Ten flights, each failing 10% of the time. How many failures are expected on average?
3. Mean endurance is 25 minutes with SD 2 minutes. What is the z-score of 21 minutes?
4. For normal data, about what percentage lies outside $\pm 2\sigma$?
5. Data has $\sigma =6$ ms. What is the approximate standard deviation of the mean of a sample of 36?