intermediate 22 minutes

Binomial Distribution

Master the binomial distribution for success/failure experiments. Learn the binomial formula, mean, variance, and applications.

binomial Bernoulli probability distribution trials
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Bernoulli Trials

A Bernoulli trial is an experiment with exactly two outcomes: success or failure.

Bernoulli Trial Examples
  • Coin flip: Heads (success) or Tails (failure)
  • Quality check: Pass (success) or Fail (failure)
  • Medical test: Positive or Negative
  • Free throw: Made or Missed

Bernoulli Distribution

Bernoulli Distribution

For a single trial with success probability p:

P(X=1)=pP(X = 1) = p P(X=0)=1−p=qP(X = 0) = 1 - p = q

Mean: μ=p\mu = p

Variance: σ2=p(1−p)=pq\sigma^2 = p(1-p) = pq

The Binomial Distribution

The binomial distribution models the number of successes in n independent Bernoulli trials.

Requirements for Binomial

The Binomial Formula

Binomial PMF

P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

Where:

  • n = number of trials
  • k = number of successes
  • p = probability of success on each trial
  • (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!} = “n choose k”
Coin Flips

Problem: Flip a fair coin 5 times. What’s P(exactly 3 heads)?

  • n = 5 trials
  • k = 3 successes (heads)
  • p = 0.5

P(X=3)=(53)(0.5)3(0.5)2P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2

=10×0.125×0.25=10×0.03125= 10 \times 0.125 \times 0.25 = 10 \times 0.03125

=0.3125= \textbf{0.3125} or 31.25%

Quality Control

Problem: A factory produces items with 10% defect rate. In a batch of 8 items, what’s P(exactly 2 are defective)?

  • n = 8 items
  • k = 2 defects
  • p = 0.10

P(X=2)=(82)(0.10)2(0.90)6P(X = 2) = \binom{8}{2} (0.10)^2 (0.90)^6

=28×0.01×0.531441= 28 \times 0.01 \times 0.531441

=0.149= \textbf{0.149} or about 15%

Cumulative Probabilities

Often we need P(X ≤ k) or P(X ≥ k), not just P(X = k).

Cumulative Probabilities

P(X≤k)=∑i=0kP(X=i)P(X \leq k) = \sum_{i=0}^{k} P(X = i)

P(X≥k)=1−P(X≤k−1)P(X \geq k) = 1 - P(X \leq k-1)

P(a≤X≤b)=P(X≤b)−P(X≤a−1)P(a \leq X \leq b) = P(X \leq b) - P(X \leq a-1)

At Least / At Most

Problem: Roll a die 4 times. p(6) = 1/6 each roll. What’s P(at least one 6)?

Method 1: Complement P(at least one 6) = 1 - P(no sixes) = 1 - P(X = 0) = 1 - (40)(1/6)0(5/6)4\binom{4}{0}(1/6)^0(5/6)^4 = 1 - 1 × 1 × 0.482 = 0.518 or about 52%

Method 2: Sum all possibilities P(X ≥ 1) = P(X=1) + P(X=2) + P(X=3) + P(X=4) (More tedious, same answer)

Mean and Variance

Binomial Mean and Variance

Mean (expected value): μ=E(X)=np\mu = E(X) = np

Variance: σ2=Var(X)=np(1−p)=npq\sigma^2 = Var(X) = np(1-p) = npq

Standard deviation: σ=np(1−p)\sigma = \sqrt{np(1-p)}

Mean and Variance

Problem: 100 free throws with 75% success rate.

  • n = 100
  • p = 0.75

Expected makes: μ = 100 × 0.75 = 75

Variance: σ² = 100 × 0.75 × 0.25 = 18.75

Standard deviation: σ = √18.75 ≈ 4.33

Interpretation: On average, 75 makes with SD of about 4.33.

Shape of Binomial Distribution

ConditionShape
p = 0.5Symmetric
p < 0.5Right-skewed
p > 0.5Left-skewed
Large nApproximately normal

When to Use Binomial

Common Mistakes

MistakeCorrection
Forgetting (nk)\binom{n}{k}Always include the combination term
Using wrong pIdentify what “success” means
Assuming independenceCheck if sampling with replacement
Wrong formula for ≥P(X ≥ k) = 1 - P(X ≤ k-1), not 1 - P(X ≤ k)

Summary

In this lesson, you learned:

  • Bernoulli trial: Single experiment with success (p) or failure (1-p)
  • Binomial distribution: Number of successes in n independent trials
  • Binomial formula: P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k}p^k(1-p)^{n-k}
  • BINS conditions: Binary, Independent, Number fixed, Same probability
  • Mean: μ = np
  • Variance: σ² = np(1-p)
  • Use complement rule for “at least one” problems

Practice Problems

1. A coin with P(heads) = 0.6 is flipped 10 times. a) What’s P(exactly 6 heads)? b) What’s E(X) and SD(X)?

2. A multiple choice test has 5 questions with 4 choices each. If you guess randomly, what’s P(getting at least 2 correct)?

3. A basketball player makes 80% of free throws. In 20 attempts, what’s the expected number of makes? What’s the standard deviation?

4. Why can’t we use binomial for drawing 5 cards from a deck and counting hearts (without replacement)?

Click to see answers

1a. n = 10, k = 6, p = 0.6 P(X=6)=(106)(0.6)6(0.4)4P(X = 6) = \binom{10}{6}(0.6)^6(0.4)^4 = 210 × 0.0467 × 0.0256 = 0.251 (about 25%)

1b. E(X) = np = 10 × 0.6 = 6 SD(X) = √(np(1-p)) = √(10 × 0.6 × 0.4) = √2.4 ≈ 1.55

2. n = 5, p = 0.25 (1 correct out of 4 choices)

P(X ≥ 2) = 1 - P(X ≤ 1) = 1 - [P(X=0) + P(X=1)]

P(X=0) = (50)(0.25)0(0.75)5\binom{5}{0}(0.25)^0(0.75)^5 = 0.2373 P(X=1) = (51)(0.25)1(0.75)4\binom{5}{1}(0.25)^1(0.75)^4 = 0.3955

P(X ≥ 2) = 1 - (0.2373 + 0.3955) = 0.367 (about 37%)

3. n = 20, p = 0.80

Expected makes: μ = 20 × 0.80 = 16 SD: σ = √(20 × 0.80 × 0.20) = √3.2 ≈ 1.79

4. Without replacement, trials are not independent.

After drawing the first card, the probability of hearts changes:

  • If first card was a heart: 12/51
  • If first card was not a heart: 13/51

The probability is not constant across trials.

(For very large populations, this effect is negligible and binomial can approximate.)

Next Steps

Explore more probability distributions:

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