Overview
The binomial distribution models the number of successes in a fixed number of independent trials, where each trial has only two outcomes (success/failure).
Conditions (BINS)
B inary: Two outcomes (success/failure)I ndependent: Trials are independentN umber: Fixed number of trials (n n n S uccess: Same probability of success (p p p
Parameters
n n n p p p q = 1 − p q = 1 - p q = 1 − p
Probability Mass Function
The probability of exactly k k k n n n
P ( X = k ) = ( n k ) p k ( 1 − p ) n − k P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} P ( X = k ) = ( k n ) p k ( 1 − p ) n − k = n ! k ! ( n − k ) ! p k q n − k = \frac{n!}{k!(n-k)!} p^k q^{n-k} = k ! ( n − k )! n ! p k q n − k Where:
( n k ) \binom{n}{k} ( k n ) k = 0 , 1 , 2 , … , n k = 0, 1, 2, \ldots, n k = 0 , 1 , 2 , … , n
Key Formulas
Mean (Expected Value)
μ = E ( X ) = n p \mu = E(X) = np μ = E ( X ) = n p Variance
σ 2 = Var ( X ) = n p ( 1 − p ) = n p q \sigma^2 = \text{Var}(X) = np(1-p) = npq σ 2 = Var ( X ) = n p ( 1 − p ) = n pq Standard Deviation
σ = n p q \sigma = \sqrt{npq} σ = n pq Notation
X ∼ Binomial ( n , p ) X \sim \text{Binomial}(n, p) X ∼ Binomial ( n , p ) or
X ∼ B ( n , p ) X \sim B(n, p) X ∼ B ( n , p ) Cumulative Probability
P ( X ≤ k ) = ∑ i = 0 k ( n i ) p i ( 1 − p ) n − i P(X \leq k) = \sum_{i=0}^{k} \binom{n}{i} p^i (1-p)^{n-i} P ( X ≤ k ) = i = 0 ∑ k ( i n ) p i ( 1 − p ) n − i Using Complement
P ( X ≥ k ) = 1 − P ( X ≤ k − 1 ) P(X \geq k) = 1 - P(X \leq k-1) P ( X ≥ k ) = 1 − P ( X ≤ k − 1 ) P ( X > k ) = 1 − P ( X ≤ k ) P(X > k) = 1 - P(X \leq k) P ( X > k ) = 1 − P ( X ≤ k ) Examples
Example 1: Basic Calculation
Fair coin flipped 5 times. P ( exactly 3 heads ) P(\text{exactly 3 heads}) P ( exactly 3 heads )
n = 5 , p = 0.5 , k = 3 n = 5, \quad p = 0.5, \quad k = 3 n = 5 , p = 0.5 , k = 3 P ( X = 3 ) = ( 5 3 ) ( 0.5 ) 3 ( 0.5 ) 2 = 10 × 0.125 × 0.25 = 0.3125 P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2 = 10 \times 0.125 \times 0.25 = 0.3125 P ( X = 3 ) = ( 3 5 ) ( 0.5 ) 3 ( 0.5 ) 2 = 10 × 0.125 × 0.25 = 0.3125 Example 2: Quality Control
A machine produces items with 10% defective rate. In a batch of 20 items:
n = 20 , p = 0.10 n = 20, \quad p = 0.10 n = 20 , p = 0.10 E ( defective ) = 20 × 0.10 = 2 items E(\text{defective}) = 20 \times 0.10 = 2 \text{ items} E ( defective ) = 20 × 0.10 = 2 items S D = 20 × 0.10 × 0.90 = 1.8 = 1.34 items SD = \sqrt{20 \times 0.10 \times 0.90} = \sqrt{1.8} = 1.34 \text{ items} S D = 20 × 0.10 × 0.90 = 1.8 = 1.34 items P ( exactly 0 defective ) P(\text{exactly 0 defective}) P ( exactly 0 defective )
P ( X = 0 ) = ( 20 0 ) ( 0.10 ) 0 ( 0.90 ) 20 = 1 × 1 × 0.1216 = 0.1216 P(X = 0) = \binom{20}{0} (0.10)^0 (0.90)^{20} = 1 \times 1 \times 0.1216 = 0.1216 P ( X = 0 ) = ( 0 20 ) ( 0.10 ) 0 ( 0.90 ) 20 = 1 × 1 × 0.1216 = 0.1216 Example 3: At Least One Success
P ( at least 1 six in 4 die rolls ) P(\text{at least 1 six in 4 die rolls}) P ( at least 1 six in 4 die rolls )
n = 4 , p = 1 6 n = 4, \quad p = \frac{1}{6} n = 4 , p = 6 1 P ( X ≥ 1 ) = 1 − P ( X = 0 ) = 1 − ( 4 0 ) ( 1 6 ) 0 ( 5 6 ) 4 P(X \geq 1) = 1 - P(X = 0) = 1 - \binom{4}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^4 P ( X ≥ 1 ) = 1 − P ( X = 0 ) = 1 − ( 0 4 ) ( 6 1 ) 0 ( 6 5 ) 4 = 1 − ( 5 6 ) 4 = 1 − 0.482 = 0.518 = 1 - \left(\frac{5}{6}\right)^4 = 1 - 0.482 = 0.518 = 1 − ( 6 5 ) 4 = 1 − 0.482 = 0.518 Example 4: Multiple Values
90% success rate on free throws. In 10 attempts, P ( 8 or more ) P(8 \text{ or more}) P ( 8 or more )
P ( X ≥ 8 ) = P ( X = 8 ) + P ( X = 9 ) + P ( X = 10 ) P(X \geq 8) = P(X=8) + P(X=9) + P(X=10) P ( X ≥ 8 ) = P ( X = 8 ) + P ( X = 9 ) + P ( X = 10 ) P ( 8 ) = ( 10 8 ) ( 0.9 ) 8 ( 0.1 ) 2 = 45 × 0.4305 × 0.01 = 0.1937 P(8) = \binom{10}{8}(0.9)^8(0.1)^2 = 45 \times 0.4305 \times 0.01 = 0.1937 P ( 8 ) = ( 8 10 ) ( 0.9 ) 8 ( 0.1 ) 2 = 45 × 0.4305 × 0.01 = 0.1937 P ( 9 ) = ( 10 9 ) ( 0.9 ) 9 ( 0.1 ) 1 = 10 × 0.3874 × 0.1 = 0.3874 P(9) = \binom{10}{9}(0.9)^9(0.1)^1 = 10 \times 0.3874 \times 0.1 = 0.3874 P ( 9 ) = ( 9 10 ) ( 0.9 ) 9 ( 0.1 ) 1 = 10 × 0.3874 × 0.1 = 0.3874 P ( 10 ) = ( 10 10 ) ( 0.9 ) 10 = 0.3487 P(10) = \binom{10}{10}(0.9)^{10} = 0.3487 P ( 10 ) = ( 10 10 ) ( 0.9 ) 10 = 0.3487 P ( X ≥ 8 ) = 0.1937 + 0.3874 + 0.3487 = 0.9298 P(X \geq 8) = 0.1937 + 0.3874 + 0.3487 = 0.9298 P ( X ≥ 8 ) = 0.1937 + 0.3874 + 0.3487 = 0.9298 Normal Approximation
When n n n
X ∼ N ( n p , n p q ) X \sim N(np, \sqrt{npq}) X ∼ N ( n p , n pq ) Rule of thumb: Use when n p ≥ 10 np \geq 10 n p ≥ 10 n ( 1 − p ) ≥ 10 n(1-p) \geq 10 n ( 1 − p ) ≥ 10
With continuity correction:
P ( X = k ) ≈ P ( k − 0.5 < Y < k + 0.5 ) P(X = k) \approx P(k - 0.5 < Y < k + 0.5) P ( X = k ) ≈ P ( k − 0.5 < Y < k + 0.5 ) where Y ∼ N ( n p , n p q ) Y \sim N(np, \sqrt{npq}) Y ∼ N ( n p , n pq )