Bernoulli & Binomial Distributions

Counting Successes

A Bernoulli($p$) trial has two outcomes: success (probability $p$) and failure (probability $1-p$).

A Binomial($n$, $p$) random variable counts successes in $n$ independent Bernoulli trials:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

$E[X] = np, \qquad \text{Var}(X) = np(1-p)$

The $\binom{n}{k}$ factor counts the number of ways to arrange $k$ successes in $n$ positions.

import math

def binom_pmf(n, p, k):
    return math.comb(n, k) * p**k * (1 - p)**(n - k)

def binom_cdf(n, p, k):
    return sum(binom_pmf(n, p, i) for i in range(k + 1))

# P(exactly 3 heads in 10 fair flips)
print(round(binom_pmf(10, 0.5, 3), 4))   # 0.1172
print(round(binom_cdf(10, 0.5, 3), 4))   # 0.1719

Why This Distribution Is Everywhere

• Quality control: number of defective items in a batch
• Clinical trials: number of patients who respond to treatment
• Surveys: number of yes responses out of $n$ respondents
• A/B testing: number of conversions out of $n$ visitors

Your Task

Implement binomial_full(n, p, k) that prints PMF$(k)$, CDF$(k)$, $E[X]$, and $\text{Var}(X)$, each rounded to 4 decimal places.