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Binomial Distributions

Key Terms

X ~ B(n, p) counts the number of successes in n independent Bernoulli(p) trials
The BINS conditions must hold: Binary outcomes, Independent trials, fixed Number of trials, constant Success probability p
P(X = k) = nCk pk(1 − p)n − k for k = 0, 1, 2, …, n
E(X) = np
 —  each trial contributes p to the expected count
Var(X) = np(1 − p)
SD(X) = √(np(1 − p))
Cumulative: P(X ≤ k) = P(X=0) + P(X=1) + … + P(X=k);   P(X ≥ k) = 1 − P(X ≤ k − 1)
Binomial Distribution X ~ B(n, p):
P(X = k) = nCk pk(1 − p)n − k,   k = 0, 1, 2, …, n

Mean: E(X) = np
Variance: Var(X) = np(1 − p)
Standard Deviation: SD(X) = √(np(1 − p))

Complement rule: P(X ≥ k) = 1 − P(X ≤ k − 1)
Combinations: nCk = n! / (k!(n − k)!)
Hot Tip: Before applying the Binomial formula, always verify the BINS conditions: Binary outcomes (only success/failure), Independent trials, fixed Number of trials n, constant Success probability p. If any condition fails, the Binomial model does not apply. For cumulative probabilities P(X ≥ k), use the complement: 1 − P(X ≤ k − 1).
Worked Example 1: X ~ B(5, 0.4). Find P(X = 3).

P(X = 3) = 5C3 × (0.4)3 × (0.6)2
= 10 × 0.064 × 0.36
= 10 × 0.02304
= 0.2304
Worked Example 2: X ~ B(10, 0.3). Find E(X), Var(X), and SD(X).

E(X) = np = 10 × 0.3 = 3
Var(X) = np(1 − p) = 10 × 0.3 × 0.7 = 2.1
SD(X) = √2.1 ≈ 1.449

From Bernoulli to Binomial

In the previous lesson, we studied Bernoulli trials — single experiments with exactly two outcomes (success and failure) and a fixed probability p of success. The Binomial distribution extends this idea: instead of one trial, we perform n independent Bernoulli trials and count the total number of successes.

We write X ~ B(n, p) to mean “X follows a Binomial distribution with n trials and success probability p.” X can take integer values from 0 (no successes) to n (all successes).

The BINS Conditions

The Binomial model is only valid when all four BINS conditions are satisfied:

  • Binary: each trial has exactly two outcomes (success or failure)
  • Independent: the outcome of one trial does not affect any other trial
  • Number: there is a fixed, predetermined number of trials n
  • Success probability: p is constant across all trials

A classic example: flipping a fair coin 10 times and counting heads. Each flip is binary (H or T), independent, n = 10 is fixed, and p = 0.5 is constant. Contrast this with drawing cards without replacement — p changes with each draw, violating the “S” condition.

The Binomial Probability Formula

To find P(X = k), the probability of exactly k successes:

P(X = k) = nCk × pk × (1 − p)n − k

This formula has two parts multiplied together:

  • nCk counts the number of ways to arrange k successes among n trials
  • pk(1 − p)n − k is the probability of any one specific arrangement of k successes and (n − k) failures

For example, P(X = 2) for n = 4, p = 0.3: there are 4C2 = 6 ways to get 2 successes in 4 trials, and each arrangement has probability (0.3)²(0.7)² = 0.0441. So P(X = 2) = 6 × 0.0441 = 0.2646.

Mean and Variance

Since X = X1 + X2 + ··· + Xn is the sum of n independent Bernoulli(p) variables, and since E and Var are additive for independent variables:

E(X) = nE(Xi) = n × p = np

Var(X) = nVar(Xi) = n × p(1 − p) = np(1 − p)

These formulas are elegant: the mean is simply the expected number of successes per trial (p) multiplied by the number of trials (n).

Cumulative Probabilities

Problems often ask for P(X ≤ k) or P(X ≥ k). For small n, calculate and sum the individual probabilities. For P(X ≥ k), use the complement:

P(X ≥ k) = 1 − P(X ≤ k − 1)

This avoids summing many terms. For example, P(X ≥ 3) = 1 − [P(X=0) + P(X=1) + P(X=2)].

Exam strategy: Always show the formula with values substituted in one clear line before evaluating. Write nCk × pk × (1−p)n−k with numbers filled in. For cumulative questions, clearly state whether you are finding P(X ≤ k) or using the complement rule. Set your working out in steps: identify n and p, state the formula, substitute, evaluate.

Mastery Practice

  1. A multiple-choice test has 8 questions, each with 4 options (one correct). A student guesses every answer randomly. Identify n and p for a Binomial model and verify the BINS conditions. Fluency

  2. X ~ B(6, 0.5). Find P(X = 4). Fluency

  3. X ~ B(5, 0.4). Find P(X = 2). Fluency

  4. X ~ B(4, 0.3). Find P(X ≤ 2) by calculating P(X = 0) + P(X = 1) + P(X = 2). Fluency

  5. X ~ B(4, 0.3). Find P(X ≥ 3) using the complement rule. Understanding

  6. X ~ B(12, 0.25). Find E(X) and Var(X). Understanding

  7. X ~ B(20, 0.4). Find SD(X) correct to 3 decimal places. Understanding

  8. For X ~ B(n, p), you are given that E(X) = 6 and Var(X) = 2.4. Find n and p. Understanding

  9. A fair coin is tossed 10 times. Let X = number of heads. Complete a partial distribution table: find P(X = 0), P(X = 1), P(X = 10), and state E(X). Understanding

  10. A factory produces light globes; 8% are defective. A quality inspector randomly tests 15 globes. Let X = number of defective globes. Find P(X = 0) and P(X ≥ 2). Interpret your answers. Problem Solving

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