Binomial Probability Mass Function

Probability and Statistics

Easy

Problem

Theory

The Binomial distribution models the number of successes in a fixed number of independent trials, where each trial has the same probability of success.

It answers the question: "If I repeat an experiment n times, each with success probability p, what is the probability of getting exactly k successes?"

The Four Conditions

A random variable follows a Binomial distribution if and only if:

1. Fixed number of trials (n)

The experiment is repeated a predetermined number of times.

2. Independent trials

The outcome of one trial does not affect the others.

3. Two outcomes per trial

Each trial results in either "success" or "failure."

4. Constant probability (p)

The probability of success is the same for every trial.

If any condition is violated, the Binomial model may not apply.

Notation and Parameters

We write X \sim \text{Binomial}(n, p) to indicate X follows a Binomial distribution with:

• n = number of trials (positive integer)
• p = probability of success on each trial ($ \leq p \leq 1$)

The random variable X can take values 0, 1, 2, ..., n.

The Probability Mass Function (PMF)

The PMF gives the probability of exactly k successes:

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

where k \in \{0, 1, 2, ..., n\}.

Components explained:

• \binom{n}{k} = \frac{n!}{k!(n-k)!} is the binomial coefficient, counting the number of ways to choose which k trials are successes
• p^k is the probability of k successes
• (1-p)^{n-k} is the probability of n-k failures

Understanding the Binomial Coefficient

The term \binom{n}{k} counts the number of ways to arrange k successes among n trials.

Example: For n = 4 trials and k = 2 successes:

\binom{4}{2} = \frac{4!}{2! \cdot 2!} = \frac{24}{2 \cdot 2} = 6

The 6 possible patterns are: SSFF, SFSF, SFFS, FSSF, FSFS, FFSS

Each pattern has probability p^2(1-p)^2, and there are 6 of them.

Worked Example: Coin Flips

Setup: Flip a fair coin (p = 0.5) 5 times. What is the probability of exactly 3 heads?

Solution:

n = 5, k = 3, p = 0.5

P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2

= \frac{5!}{3! \cdot 2!} \cdot 0.125 \cdot 0.25

= 10 \cdot 0.03125 = 0.3125

There is a 31.25% chance of getting exactly 3 heads.

Computing the Full PMF

Setup: n = 4, p = 0.3

For each value of k:

P(X = 0) = \binom{4}{0}(0.3)^0(0.7)^4 = 1 \cdot 1 \cdot 0.2401 = 0.2401

P(X = 1) = \binom{4}{1}(0.3)^1(0.7)^3 = 4 \cdot 0.3 \cdot 0.343 = 0.4116

P(X = 2) = \binom{4}{2}(0.3)^2(0.7)^2 = 6 \cdot 0.09 \cdot 0.49 = 0.2646

P(X = 3) = \binom{4}{3}(0.3)^3(0.7)^1 = 4 \cdot 0.027 \cdot 0.7 = 0.0756

P(X = 4) = \binom{4}{4}(0.3)^4(0.7)^0 = 1 \cdot 0.0081 \cdot 1 = 0.0081

Verification: 0.2401 + 0.4116 + 0.2646 + 0.0756 + 0.0081 = 1.0 ✓

The Cumulative Distribution Function (CDF)

The CDF gives the probability of at most k successes:

F(k) = P(X \leq k) = \sum_{i=0}^{k} \binom{n}{i} p^i (1-p)^{n-i}

Properties:

• F(k) is a step function, increasing at each integer
• F(-1) = 0 (by convention)
• F(n) = 1

Computing CDF Values

Setup: n = 4, p = 0.3 (continuing previous example)

F(0) = P(X \leq 0) = 0.2401

F(1) = P(X \leq 1) = 0.2401 + 0.4116 = 0.6517

F(2) = P(X \leq 2) = 0.6517 + 0.2646 = 0.9163

F(3) = P(X \leq 3) = 0.9163 + 0.0756 = 0.9919

F(4) = P(X \leq 4) = 0.9919 + 0.0081 = 1.0

Using the CDF for Range Probabilities

The CDF makes it easy to compute probabilities over ranges:

P(X > k):

P(X > k) = 1 - F(k) = 1 - P(X \leq k)

P(X \geq k):

P(X \geq k) = 1 - F(k-1) = 1 - P(X \leq k-1)

P(a \leq X \leq b):

P(a \leq X \leq b) = F(b) - F(a-1)

Worked Example: Range Probability

Setup: n = 10, p = 0.4. Find P(3 \leq X \leq 6).

P(3 \leq X \leq 6) = F(6) - F(2)

= P(X \leq 6) - P(X \leq 2)

= \sum_{k=0}^{6} \binom{10}{k}(0.4)^k(0.6)^{10-k} - \sum_{k=0}^{2} \binom{10}{k}(0.4)^k(0.6)^{10-k}

Computing: F(6) \approx 0.9452 and F(2) \approx 0.1673

P(3 \leq X \leq 6) \approx 0.9452 - 0.1673 = 0.7779

Expected Value (Mean)

The expected number of successes is:

E[X] = np

Derivation:

X = X_1 + X_2 + ... + X_n where each X_i \sim \text{Bernoulli}(p)

By linearity of expectation:

E[X] = E[X_1] + E[X_2] + ... + E[X_n] = p + p + ... + p = np

Example: In 100 coin flips with p = 0.5, we expect 100 \times 0.5 = 50 heads.

Variance

The variance of the number of successes is:

\text{Var}(X) = np(1-p)

Derivation:

Since X_1, X_2, ..., X_n are independent:

\text{Var}(X) = \text{Var}(X_1) + ... + \text{Var}(X_n) = np(1-p)

Standard deviation:

\sigma = \sqrt{np(1-p)}

Example: For n = 100, p = 0.5:

\text{Var}(X) = 100 \times 0.5 \times 0.5 = 25

\sigma = \sqrt{25} = 5

Mode of the Distribution

The mode (most likely value) is approximately:

\text{mode} \approx \lfloor (n+1)p \rfloor \text{ or } \lceil (n+1)p \rceil - 1

For integer (n+1)p, there may be two modes.

Example: n = 10, p = 0.3

(n+1)p = 11 \times 0.3 = 3.3

Mode = \lfloor 3.3 \rfloor = 3

The most likely number of successes is 3.

Shape of the Distribution

The shape depends on p:

p < 0.5: Right-skewed (tail extends toward larger values)

p = 0.5: Symmetric

p > 0.5: Left-skewed (tail extends toward smaller values)

As n increases, the distribution becomes more symmetric and bell-shaped (by the Central Limit Theorem).

Normal Approximation

For large n, the Binomial can be approximated by a Normal distribution:

X \approx N(np, np(1-p))

Rule of thumb: The approximation is reasonable when:

• np \geq 10 and n(1-p) \geq 10

Continuity correction: Since Binomial is discrete and Normal is continuous:

P(X \leq k) \approx \Phi\left(\frac{k + 0.5 - np}{\sqrt{np(1-p)}}\right)

where \Phi is the standard Normal CDF.

Poisson Approximation

For large n and small p (with np = \lambda moderate):

\text{Binomial}(n, p) \approx \text{Poisson}(\lambda = np)

Rule of thumb: Works well when n \geq 20 and p \leq 0.05.

This is useful because Poisson probabilities are easier to compute.

Relationship to Bernoulli

The Binomial distribution is the sum of independent Bernoulli trials:

X = \sum_{i=1}^{n} X_i

where X_i \sim \text{Bernoulli}(p) are independent.

Special case: \text{Binomial}(1, p) = \text{Bernoulli}(p)

Sum of Binomials

If X \sim \text{Binomial}(n_1, p) and Y \sim \text{Binomial}(n_2, p) are independent with the same p:

X + Y \sim \text{Binomial}(n_1 + n_2, p)

This property does not hold if p values differ.

Applications in Machine Learning

A/B testing:

Comparing conversion rates between two versions. Each user's action is Bernoulli, total conversions are Binomial.

Classification metrics:

Number of true positives in n predictions follows Binomial if predictions are independent.

Bootstrap sampling:

Number of times a specific sample appears follows approximately \text{Binomial}(n, 1/n) \approx \text{Poisson}(1).

Dropout regularization:

Number of neurons kept in a layer follows Binomial distribution.

Maximum Likelihood Estimation

Given observations from \text{Binomial}(n, p) with known n, the MLE for p is:

\hat{p} = \frac{\bar{x}}{n}

where \bar{x} is the sample mean of observed counts.

If you observe a single value x:

\hat{p} = \frac{x}{n}

Properties Summary

• Support: \{0, 1, 2, ..., n\}
• Parameters: n \in \{1, 2, ...\}, p \in [0, 1]
• PMF: P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}
• Mean: E[X] = np
• Variance: \text{Var}(X) = np(1-p)
• Skewness: \frac{1-2p}{\sqrt{np(1-p)}}
• MGF: M_X(t) = [(1-p) + pe^t]^n

Examples

Hints

1. Use math.comb(n, i) for each binomial coefficient.
2. Build probabilities for i from 0 through k, then use the last value as the PMF and their sum as the CDF.

Requirements

• Use the stated binomial PMF
• Sum success counts from 0 through k for the CDF
• Return exactly pmf and cdf in a dictionary of Python floats

Constraints

• n is an integer between 1 and 100
• k is an integer between 0 and n
• p is between 0 and 1
• Use the Python standard library only

Starter Code

import math

def binomial_pmf_cdf(n: int, p: float, k: int) -> dict:
    """
    Returns a dictionary with pmf and cdf.
    """
    # Write code here
    pass

Test Cases