Loading…
Apply the binomial distribution to count successes in fixed trials with conditions BINS.
Learn step-by-step with practice exercises built right in.
Random variable ~ Binomial if:
A quiz has 10 multiple choice questions, each with 4 choices. A student guesses randomly. What is the probability of exactly 3 correct answers?
This is binomial with n=10, p=0.25 (one correct out of 4), k=3. P(X=3) = C(10,3) × (0.25)³ × (0.75)⁷ = 120 × 0.0156 × 0.1335 ≈ 0.250. About 25% chance of exactly 3 correct by random guessing on a 10-question quiz.
A vaccine is 85% effective. In a sample of 20 vaccinated individuals, find the probability that at least 18 are protected (X ≥ 18).
Avoid these 3 frequent errors
Review key concepts with our flashcard system
Explore more AP Statistics topics
Example: Draw 10 cards with replacement, count red cards. ~ Binomial(10, 0.5)
NOT binomial: Drawing without replacement (probabilities change)
Where:
Scenario: 80% of AP Stats students pass the exam. If 5 students take the exam, what's students pass?
Check: or 41%
What's ?
Example: 5 students,
If large enough ( AND ), then:
Use normal curve to approximate binomial probabilities (easier than formula!)
Continuity correction: For exact binomial, ≈ using normal
Two outcomes, fixed , independent trials, constant ? → Binomial Need ? → Use formula Need ? → Sum probabilities or use table/calculator
Always state BINS conditions explicitly: "This is binomial because..." Common calculation: " means exactly successes"; " means at least successes."
Binomial with n=20, p=0.85. Find P(X≥18) = P(X=18) + P(X=19) + P(X=20). P(X=18) = C(20,18) × (0.85)¹⁸ × (0.15)² ≈ 190 × 0.0394 × 0.0225 ≈ 0.169. P(X=19) = C(20,19) × (0.85)¹⁹ × (0.15)¹ ≈ 20 × 0.0335 × 0.15 ≈ 0.100. P(X=20) = (0.85)²⁰ ≈ 0.0388. P(X≥18) ≈ 0.169 + 0.100 + 0.039 ≈ 0.308. About 31% chance that 18 or more are protected.
Manufacturing produces items with defect rate p=0.02. In a batch of 50, find the probability of at most 2 defects. Also calculate E(X) and SD(X).
Binomial n=50, p=0.02. P(X≤2) = P(X=0) + P(X=1) + P(X=2). P(X=0) = C(50,0)(0.02)⁰(0.98)⁵⁰ ≈ 0.364. P(X=1) = C(50,1)(0.02)¹(0.98)⁴⁹ ≈ 50 × 0.02 × 0.371 ≈ 0.371. P(X=2) = C(50,2)(0.02)²(0.98)⁴⁸ ≈ 1225 × 0.0004 × 0.379 ≈ 0.186. P(X≤2) ≈ 0.921. E(X) = np = 50(0.02) = 1. SD(X) = √(np(1-p)) = √(50×0.02×0.98) ≈ 0.99. Over 92% probability of 2 or fewer defects; mean defects is 1 with SD ≈ 1.