Binomial Distribution

When to Use Binomial

BINS conditions:

Binary: Each trial has two outcomes (success/failure)
Independent: Trials independent of each other
Number: Fixed number of trials (n)
Same: Probability of success (p) same for each trial

If BINS met → Use Binomial distribution

Notation: X ~ Binomial(n, p)

Binomial Probability Formula

Probability of exactly k successes in n trials:

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

Where:

• $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient
• n = number of trials
• k = number of successes
• p = probability of success on each trial

Example 1: Coin Flips

Flip fair coin 5 times. Find P(exactly 3 heads).

Check BINS:

• Binary: Heads or tails ✓
• Independent: Flips independent ✓
• Number: n = 5 trials ✓
• Same: p = 0.5 each flip ✓

Calculate:

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

$= \frac{5!}{3!2!} (0.5)^3 (0.5)^2 = 10(0.125)(0.25) = 0.3125$

Example 2: Free Throws

Basketball player makes 70% of free throws. Shoots 10. Find P(exactly 8 makes).

X ~ Binomial(10, 0.7)

$P(X = 8) = \binom{10}{8} (0.7)^8 (0.3)^2$

$= 45(0.05764801)(0.09) \approx 0.2335$

Calculating Binomial Coefficient

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Calculator: nCr function

• On TI-83/84: 5 nCr 3 = 10

Example: $\binom{5}{3} = \frac{5!}{3!2!} = \frac{120}{6 \cdot 2} = 10$

Mean and Standard Deviation

Mean (Expected Value):

$\mu_X = np$

Standard Deviation:

$\sigma_X = \sqrt{np(1-p)}$

Example: n = 100 free throws, p = 0.7

$\mu_X = 100(0.7) = 70$ $\sigma_X = \sqrt{100(0.7)(0.3)} = \sqrt{21} \approx 4.58$

Interpretation: Expect about 70 makes, typically within about 4.58 of that

Cumulative Probabilities

P(X ≤ k): Use binomcdf on calculator

P(X < k): P(X ≤ k-1)

P(X ≥ k): 1 - P(X ≤ k-1)

P(X > k): 1 - P(X ≤ k)

Example: X ~ Binomial(20, 0.3), find P(X ≤ 5)

Calculator: binomcdf(20, 0.3, 5) ≈ 0.4164

Example: P(X ≥ 8) = 1 - P(X ≤ 7) = 1 - binomcdf(20, 0.3, 7) ≈ 0.0867

Calculator Commands (TI-83/84)

binompdf(n, p, k): P(X = k)

• Example: binompdf(10, 0.7, 8)

binomcdf(n, p, k): P(X ≤ k)

• Example: binomcdf(10, 0.7, 8)

Access: 2nd VARS (DISTR) → binompdf or binomcdf

Probability Distribution Graph

For Binomial(10, 0.5):

• Symmetric (when p = 0.5)
• Centered at mean (np = 5)
• Bell-shaped (approximates normal for large n)

For Binomial(10, 0.2):

• Right-skewed (when p < 0.5)
• Centered at mean (np = 2)

For Binomial(10, 0.8):

• Left-skewed (when p > 0.5)
• Centered at mean (np = 8)

Normal Approximation

When n is large, Binomial approximates Normal:

Rule of thumb: Use if np ≥ 10 and n(1-p) ≥ 10

Then: X ~ N(np, √(np(1-p))) approximately

Example: X ~ Binomial(100, 0.5)

• np = 50 ≥ 10 ✓
• n(1-p) = 50 ≥ 10 ✓
• Approximate: X ~ N(50, 5)

Use continuity correction: P(X ≤ 45) ≈ P(Y ≤ 45.5) where Y ~ N(50, 5)

Sampling Without Replacement

Technically not binomial (independence violated)

10% condition: If sample size < 10% of population, binomial is good approximation

Example: 5 cards from 52-card deck

• 5/52 ≈ 9.6% < 10%
• Can use binomial as approximation

Example: 20 cards from 52-card deck

• 20/52 ≈ 38% > 10%
• Should use hypergeometric distribution, not binomial

Common Applications

Quality control: Defective items in sample
Medical: Treatment success in patients
Testing: Correct answers by guessing
Genetics: Offspring with certain trait
Sports: Makes/misses in attempts

Example 3: Multiple-Choice Test

20 questions, 5 choices each. Find P(pass by guessing) if passing is 60%.

X ~ Binomial(20, 0.2)

Pass means X ≥ 12

$P(X \geq 12) = 1 - P(X \leq 11)$

Calculator: 1 - binomcdf(20, 0.2, 11) ≈ 0.0009

Very unlikely to pass by guessing!

Common Mistakes

❌ Forgetting to check BINS conditions
❌ Using binomial when sampling without replacement (>10% of population)
❌ Confusing P(X ≤ k) with P(X < k)
❌ Using wrong formula (mean, SD, or probability)
❌ Calculator syntax errors

Practice Strategy

1. Verify BINS: All four conditions met?
2. Identify: n = ? and p = ?
3. Determine: What are we finding? P(X = k)? P(X ≤ k)?
4. Calculate: Use formula or calculator
5. Check: Does answer make sense?

Quick Reference

BINS Conditions: Binary, Independent, Number fixed, Same probability

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

Mean: $\mu = np$

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

Calculator:

• binompdf(n, p, k) for P(X = k)
• binomcdf(n, p, k) for P(X ≤ k)

Remember: Check BINS conditions first! If met, binomial distribution provides powerful tool for calculating probabilities of success counts.