Grandeur

Math221 Project

3 min read

🟥 Slide 1: What is Binomial Distribution?

  • A binomial distribution models the number of successes in a fixed number of independent trials, where each trial has the same probability of success.

  • Formula:

    P(X=x)=(nx)⋅px⋅(1−p)n−xP(X = x) = \binom{n}{x} \cdot p^x \cdot (1-p)

Key terms:

  • n: number of trials

  • p: probability of success

  • x: number of successes

🟧 Slide 2: Why is Binomial Distribution Important?

  • It helps us understand and calculate the likelihood of outcomes when there’s a clear pass/fail or yes/no situation.

  • It’s widely used because many real-life events can be broken into binary outcomes: win/lose, click/no click, success/failure, etc.

🟨 Slide 3: Real-Life Applications

Examples:

  • Manufacturing: What are the chances that exactly 8 out of 10 products are defect-free?

  • Healthcare: What’s the probability 4 out of 5 patients respond to a treatment?

  • Marketing: How likely is it that 30% of recipients click a link in an email campaign?

  • Education: Modeling the number of students who pass a standardized test if the pass rate is known

🟩 Slide 4: Scenario Setup for Problems

Scenario:
A phone factory produces smartphones. Each phone has a 90% chance of passing inspection. An inspector tests 10 phones per batch.

Let X∼Binomial(n=10,p=0.9)X \sim \text{Binomial}(n = 10, p = 0.9)X∼Binomial(n=10,p=0.9)

We’ll now solve 4 problems based on this.

🟦 Slide 5: Problem 1: Exactly 9 Phones Pass

Question:

What is the probability that exactly 9 out of 10 phones pass inspection?

Solution:

P(X=9)=(109)⋅(0.9)9⋅(0.1)1=10⋅0.3874⋅0.1≈0.3874P(X = 9) = \binom{10}{9} \cdot (0.9)^9 \cdot (0.1)^1 = 10 \cdot 0.3874 \cdot 0.1 ≈ \boxed{0.3874}P(X=9)=(910​)⋅(0.9)9⋅(0.1)1=10⋅0.3874⋅0.1≈0.3874​

🟪 Slide 6: Problem 2: At Most 8 Pass

Question:

What is the probability that at most 8 phones pass?

P(X≤8)=1−P(9)−P(10)P(X \leq 8) = 1 - P(9) - P(10)P(X≤8)=1−P(9)−P(10)

We already have:

  • P(9)≈0.3874P(9) ≈ 0.3874P(9)≈0.3874

  • P(10)=(1010)(0.9)10≈0.3487P(10) = \binom{10}{10}(0.9)^{10} ≈ 0.3487P(10)=(1010​)(0.9)10≈0.3487

So:

P(X≤8)=1−(0.3874+0.3487)=0.2639P(X \leq 8) = 1 - (0.3874 + 0.3487) = \boxed{0.2639}P(X≤8)=1−(0.3874+0.3487)=0.2639​

🟥 Slide 7: Problem 3: At Least 1 Phone Fails

Question:

What is the probability that at least one phone fails inspection?

Shortcut:

P(at least 1 fail)=1−P(all pass)=1−(0.9)10≈1−0.3487=0.6513P(\text{at least 1 fail}) = 1 - P(\text{all pass}) = 1 - (0.9)^{10} ≈ 1 - 0.3487 = \boxed{0.6513}P(at least 1 fail)=1−P(all pass)=1−(0.9)10≈1−0.3487=0.6513​

🟧 Slide 8: Problem 4: Batch Rejected If < 8 Pass

Question:

What’s the probability the batch is rejected (fewer than 8 pass)?

P(X<8)=P(0)+P(1)+⋯+P(7)P(X < 8) = P(0) + P(1) + \dots + P(7)P(X<8)=P(0)+P(1)+⋯+P(7)

Instead of computing all, we shortcut:

P(X<8)=1−P(8)−P(9)−P(10)P(X < 8) = 1 - P(8) - P(9) - P(10)P(X<8)=1−P(8)−P(9)−P(10)

You compute:

  • P(8)=(108)(0.9)8(0.1)2≈0.1937P(8) = \binom{10}{8}(0.9)^8(0.1)^2 ≈ 0.1937P(8)=(810​)(0.9)8(0.1)2≈0.1937

  • So:

P(X<8)=1−(0.1937+0.3874+0.3487)=0.0702P(X < 8) = 1 - (0.1937 + 0.3874 + 0.3487) = \boxed{0.0702}P(X<8)=1−(0.1937+0.3874+0.3487)=0.0702​

🟨 Slide 9: Conclusion: Why It Matters

Recap:

  • Binomial distribution is essential for predicting success/failure outcomes

  • It’s simple but powerful — used in science, business, medicine, tech

  • Mastering it helps you reason about uncertainty with confidence

Graph View