Binomial Distribution

Binomial Setting (BINS)

A random variable $X$ follows a binomial distribution if:

• Binary: Each trial has exactly two outcomes (success/failure)
• Independent: Trials are independent of each other
• Number: Fixed number of trials $n$
• Success: Same probability of success $p$ on each trial

Notation: $X \sim B(n, p)$ or $X \sim \text{Binomial}(n, p)$

Binomial Probability Formula

The 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 choose k").

Mean and Standard Deviation

$\mu_X = np$

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

Calculator Commands

• $P(X = k)$: binompdf($n$, $p$, $k$)
• $P(X \leq k)$: binomcdf($n$, $p$, $k$)
• $P(X \geq k)$: 1 − binomcdf($n$, $p$, $k-1$)

The 10% Condition

When sampling without replacement, trials are not truly independent. However, we can treat them as approximately independent if:

$n < 0.10 \cdot N$

where $N$ is the population size. This is the 10% condition.

Example

A basketball player makes 75% of free throws. In 10 attempts, what's the probability of making exactly 8?

$P(X = 8) = \binom{10}{8}(0.75)^8(0.25)^2 = 45(0.1001)(0.0625) \approx 0.2816$

Shape of Binomial Distribution

• If $p = 0.5$: symmetric
• If $p < 0.5$: skewed right
• If $p > 0.5$: skewed left
• As $n$ increases, the distribution becomes more symmetric

Normal Approximation

When $np \geq 10$ and $n(1-p) \geq 10$, the binomial distribution is approximately Normal: $X \approx N(np, \sqrt{np(1-p)})$

AP Tip: Always verify the BINS conditions before using the binomial distribution. State each condition explicitly on free-response questions.