Binomial Distribution

Binary outcomes over multiple trials

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

Verify BINS: All four conditions met?
Identify: n = ? and p = ?
Determine: What are we finding? P(X = k)? P(X ≤ k)?
Calculate: Use formula or calculator
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.

📚 Practice Problems

1Problem 1easy

❓ Question:

Does flipping a coin 10 times and counting heads follow a binomial distribution? Verify all four conditions.

💡 Show Solution

Step 1: List the four binomial conditions (BINS) B - Binary: Only two outcomes (success/failure) I - Independent: Trials are independent N - Number: Fixed number of trials S - Same probability: Probability of success stays constant

Step 2: Check Binary condition Each flip: Either Heads (success) or Tails (failure) Only TWO possible outcomes per trial ✓

Step 3: Check Independent condition Each coin flip is independent

Coin has no memory
Result of one flip doesn't affect others
P(H on flip 5 | H on flip 1) = P(H on flip 5) ✓

Step 4: Check Number (fixed) condition We flip exactly 10 times

Not random number of flips
Fixed at n = 10 ✓

Step 5: Check Same probability condition P(Heads) = 0.5 on every flip

Fair coin
Probability doesn't change from trial to trial
p = 0.5 for all flips ✓

Step 6: Conclusion ALL FOUR CONDITIONS are met!

This IS a binomial distribution: X ~ Binomial(n = 10, p = 0.5)

Where X = number of heads in 10 flips

Step 7: What would the distribution look like? X can be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 Mean: E(X) = np = 10(0.5) = 5 SD: SD(X) = √[np(1-p)] = √[10(0.5)(0.5)] = √2.5 ≈ 1.58

Answer: YES, this follows a binomial distribution with n = 10 and p = 0.5. All four BINS conditions are satisfied: Binary outcomes (H or T), Independent trials, fixed Number of trials (10), Same probability of success (p = 0.5) on each trial.

2Problem 2easy

❓ Question:

If 70% of adults own a smartphone, and you randomly select 5 adults, what is the probability that exactly 3 own a smartphone?

💡 Show Solution

Step 1: Verify binomial conditions

Binary: Own or don't own (success/failure) ✓
Independent: Random selection, assume independent ✓
Number: Fixed at n = 5 ✓
Same probability: p = 0.7 for each ✓

This is binomial!

Step 2: Identify parameters n = 5 (number of trials) p = 0.7 (probability of success) x = 3 (number of successes we want)

Find: P(X = 3)

Step 3: Use binomial probability formula P(X = x) = C(n,x) · p^x · (1-p)^(n-x)

Where C(n,x) = n!/(x!(n-x)!)

Step 4: Calculate C(5,3) C(5,3) = 5!/(3!·2!) = (5·4·3!)/(3!·2·1) = (5·4)/(2·1) = 20/2 = 10

Step 5: Calculate P(X = 3) P(X = 3) = C(5,3) · (0.7)³ · (0.3)² = 10 · (0.7)³ · (0.3)² = 10 · 0.343 · 0.09 = 10 · 0.03087 = 0.3087

Step 6: Interpret P(X = 3) ≈ 0.309 or 30.9%

This means:

About 31% chance exactly 3 out of 5 own smartphone
Most likely outcome (can verify by checking others)

Step 7: Why the formula works C(5,3) = 10: Number of ways to choose which 3 own it (0.7)³: Probability those 3 own it (0.3)²: Probability the other 2 don't own it

Example sequence: SSSFF (success, success, success, failure, failure) P(SSSFF) = 0.7 · 0.7 · 0.7 · 0.3 · 0.3 = (0.7)³(0.3)²

But there are 10 different sequences with 3 S's and 2 F's!

Answer: P(X = 3) = 0.3087 or about 30.9%

The probability that exactly 3 out of 5 randomly selected adults own a smartphone is approximately 0.309.

3Problem 3medium

❓ Question:

A multiple choice quiz has 10 questions, each with 4 choices. If you guess randomly on all questions, what is the probability you pass (get at least 6 correct)?

💡 Show Solution

Step 1: Set up as binomial n = 10 questions p = 1/4 = 0.25 (probability of guessing correctly) X = number correct

Find: P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

Step 2: Calculate P(X = 6) P(X = 6) = C(10,6) · (0.25)⁶ · (0.75)⁴

C(10,6) = 10!/(6!·4!) = 210

P(X = 6) = 210 · (0.25)⁶ · (0.75)⁴ = 210 · 0.000244 · 0.316 = 0.0162

Step 3: Calculate P(X = 7) P(X = 7) = C(10,7) · (0.25)⁷ · (0.75)³

C(10,7) = 120

P(X = 7) = 120 · (0.25)⁷ · (0.75)³ = 120 · 0.000061 · 0.422 = 0.0031

Step 4: Calculate P(X = 8) P(X = 8) = C(10,8) · (0.25)⁸ · (0.75)²

C(10,8) = 45

P(X = 8) = 45 · (0.25)⁸ · (0.75)² = 45 · 0.000015 · 0.5625 = 0.00038

Step 5: Calculate P(X = 9) P(X = 9) = C(10,9) · (0.25)⁹ · (0.75)¹

C(10,9) = 10

P(X = 9) = 10 · (0.25)⁹ · 0.75 = 10 · 0.0000038 · 0.75 = 0.000029

Step 6: Calculate P(X = 10) P(X = 10) = C(10,10) · (0.25)¹⁰ · (0.75)⁰

C(10,10) = 1

P(X = 10) = 1 · (0.25)¹⁰ · 1 = 0.00000095

Step 7: Sum all probabilities P(X ≥ 6) = 0.0162 + 0.0031 + 0.00038 + 0.000029 + 0.00000095 ≈ 0.0197

Step 8: Interpret P(pass by guessing) ≈ 0.020 = 2.0%

Very unlikely! Only about 2% chance of passing by pure guessing.

Step 9: Compare to expected value E(X) = np = 10(0.25) = 2.5 Expect to get about 2.5 questions right by guessing Passing requires 6+ correct - well above average!

Answer: P(X ≥ 6) ≈ 0.020 or 2.0%

There's only about a 2% chance of passing by randomly guessing. The expected number correct is only 2.5, far below the 6 needed to pass.

4Problem 4medium

❓ Question:

For a binomial random variable with n = 20 and p = 0.4, find the mean, variance, and standard deviation.

💡 Show Solution

Step 1: Identify given information X ~ Binomial(n = 20, p = 0.4)

n = 20 trials p = 0.4 probability of success q = 1 - p = 0.6 probability of failure

Step 2: Calculate mean (expected value) Formula: μ = E(X) = np

μ = 20 × 0.4 = 8

Step 3: Calculate variance Formula: σ² = Var(X) = np(1-p) = npq

σ² = 20 × 0.4 × 0.6 = 20 × 0.24 = 4.8

Step 4: Calculate standard deviation Formula: σ = SD(X) = √[np(1-p)] = √(npq)

σ = √4.8 = √(24/5) = (2√30)/5 ≈ 2.19

Step 5: Interpret the results Mean = 8:

On average, expect 8 successes out of 20 trials
Makes sense: 40% of 20 = 8

Variance = 4.8:

Measures spread of distribution
Typical squared distance from mean

Standard Deviation ≈ 2.19:

Typical distance from mean is about 2.19
Most values fall within μ ± 2σ = 8 ± 4.38
So typically between 3.62 and 12.38 successes

Step 6: Why these formulas? The binomial is sum of n independent Bernoulli trials: X = X₁ + X₂ + ... + X₂₀

Each Xᵢ has:

E(Xᵢ) = p
Var(Xᵢ) = p(1-p)

By properties of sums:

E(X) = E(X₁) + ... + E(X₂₀) = np
Var(X) = Var(X₁) + ... + Var(X₂₀) = np(1-p)

Step 7: Verify reasonableness If p = 0.5 (coin flip):

Mean would be 10 (half of 20) ✓
Maximum variance when p = 0.5

Our p = 0.4 is close to 0.5:

Mean = 8 (slightly less than 10) ✓
Variance = 4.8 (slightly less than maximum) ✓

Answer: Mean: μ = 8 Variance: σ² = 4.8 Standard Deviation: σ ≈ 2.19

On average, we expect 8 successes out of 20 trials, with a typical deviation of about 2.19 from this mean.

5Problem 5hard

❓ Question:

A basketball player makes 75% of free throws. In a game, she attempts 12 free throws. What is the probability she makes at least 10? Also find the probability she makes exactly the expected number.

💡 Show Solution

Step 1: Set up binomial distribution X ~ Binomial(n = 12, p = 0.75) X = number of successful free throws

Step 2: Find P(X ≥ 10) P(X ≥ 10) = P(X = 10) + P(X = 11) + P(X = 12)

Step 3: Calculate P(X = 10) P(X = 10) = C(12,10) · (0.75)¹⁰ · (0.25)²

C(12,10) = 12!/(10!·2!) = (12·11)/2 = 66

P(X = 10) = 66 · (0.75)¹⁰ · (0.25)² = 66 · 0.0563 · 0.0625 = 0.2323

Step 4: Calculate P(X = 11) P(X = 11) = C(12,11) · (0.75)¹¹ · (0.25)¹

C(12,11) = 12

P(X = 11) = 12 · (0.75)¹¹ · 0.25 = 12 · 0.0422 · 0.25 = 0.1267

Step 5: Calculate P(X = 12) P(X = 12) = C(12,12) · (0.75)¹² · (0.25)⁰

C(12,12) = 1

P(X = 12) = 1 · (0.75)¹² · 1 = 0.0317

Step 6: Sum for P(X ≥ 10) P(X ≥ 10) = 0.2323 + 0.1267 + 0.0317 = 0.3907

Step 7: Find expected value E(X) = np = 12 × 0.75 = 9

The expected number of makes is 9.

Step 8: Find P(X = 9) P(X = 9) = C(12,9) · (0.75)⁹ · (0.25)³

C(12,9) = 12!/(9!·3!) = (12·11·10)/(3·2·1) = 220

P(X = 9) = 220 · (0.75)⁹ · (0.25)³ = 220 · 0.0751 · 0.0156 = 0.2581

Step 9: Interpret results P(X ≥ 10) ≈ 0.391 or 39.1%

About 39% chance of making at least 10 out of 12
Fairly likely for a 75% shooter

P(X = 9) ≈ 0.258 or 25.8%

Most likely single outcome!
But still only 26% chance of getting EXACTLY the expected value
Other values around 9 are also likely

Step 10: Visualize distribution The distribution is skewed left (p > 0.5) Peak near x = 9 Spread: most values between 6 and 12

Answer: P(X ≥ 10) ≈ 0.391 or 39.1% E(X) = 9 P(X = 9) ≈ 0.258 or 25.8%

There's about a 39% chance she makes at least 10 free throws. While the expected number is 9, the probability of making exactly 9 is only about 26% - the distribution has considerable spread around the mean.

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