Example: Draw 10 cards with replacement, count red cards. $X$ ~ Binomial(10, 0.5)

NOT binomial: Drawing without replacement (probabilities change)

Binomial Probability Formula

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

Where:

• $n$ = number of trials
• $k$ = number of successes (0 to $n$)
• $p$ = probability of success on each trial
• $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ = number of ways to choose $k$ from $n$

Worked Example

Scenario: 80% of AP Stats students pass the exam. If 5 students take the exam, what's $P(X = 4)$ students pass?

• $n = 5$, $k = 4$, $p = 0.8$, $(1-p) = 0.2$

$P(X = 4) = \binom{5}{4} (0.8)^4 (0.2)^1 = 5 × 0.4096 × 0.2 = 0.4096$

Check: $P(X = 4) ≈ 0.41$ or 41%

What's $P(X ≥ 4) = P(X=4) + P(X=5)$?

$P(X=5) = \binom{5}{5} (0.8)^5 (0.2)^0 = 1 × 0.32768 × 1 = 0.32768$

$P(X ≥ 4) ≈ 0.41 + 0.33 ≈ 0.74$

Mean and Standard Deviation

$E(X) = \mu = np$

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

Example: 5 students, $p=0.8$

• $E(X) = 5(0.8) = 4$ students
• $SD(X) = \sqrt{5(0.8)(0.2)} = \sqrt{0.8} ≈ 0.894$

Normal Approximation to Binomial

If $n$ large enough ($np ≥ 10$ AND $n(1-p) ≥ 10$), then:

$X \text{ approximately } N(np, \sqrt{np(1-p)})$

Use normal curve to approximate binomial probabilities (easier than formula!)

Continuity correction: For exact binomial, $P(X=k)$ ≈ $P(k-0.5 < X < k+0.5)$ using normal

Common Mistakes

• Forgetting BINS conditions before applying binomial
• Using $(1-p)^{n-k}$ as $(1-p)^n$
• Confusing $P(X = k)$ with $P(X \leq k)$
• Sampling without replacement (violates independence)

Decision Rule

Two outcomes, fixed $n$, independent trials, constant $p$? → Binomial Need $P(X = k)$? → Use formula Need $P(X \leq k)$? → Sum probabilities or use table/calculator

AP Exam Tip

Always state BINS conditions explicitly: "This is binomial because..." Common calculation: "$P(X=k)$ means exactly $k$ successes"; "$P(X \geq k)$ means at least $k$ successes."