Probability

• Sample space ($S$): The set of all possible outcomes of an experiment
• Event: Any subset of the sample space
• Probability: A number between 0 and 1 that measures the likelihood of an event

1. Legitimate probability values: $0 \leq P(A) \leq 1$ for any event $A$
2. Sum of probabilities: $P(S) = 1$ (the probability of the entire sample space is 1)
3. Complement rule: $P(A^c) = 1 - P(A)$, where $A^c$ is the complement of $A$

For any two events $A$ and $B$:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

If $A$ and $B$ are mutually exclusive (disjoint, cannot both occur): $P(A \cup B) = P(A) + P(B)$

$P(A \cap B) = P(A) \cdot P(B | A)$

If $A$ and $B$ are independent: $P(A \cap B) = P(A) \cdot P(B)$

$P(B | A) = \frac{P(A \cap B)}{P(A)}$

The probability of $B$ given that $A$ has occurred. Note that $P(A)$ must be greater than 0.

Events $A$ and $B$ are independent if the occurrence of one does not affect the probability of the other:

$P(B | A) = P(B) \quad \text{and} \quad P(A | B) = P(A)$

Equivalently: $P(A \cap B) = P(A) \cdot P(B)$

Important: Independence is not the same as mutually exclusive. In fact, if two events are both mutually exclusive and both have non-zero probability, they cannot be independent.

To check independence on the AP exam:

1. Calculate $P(A)$ and $P(B)$ separately
2. Calculate $P(A \cap B)$
3. Check whether $P(A \cap B) = P(A) \cdot P(B)$

Alternatively, check whether $P(B | A) = P(B)$.

$P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B | A) \cdot P(A) + P(B | A^c) \cdot P(A^c)}$

Used to find the probability of a “cause” given an observed “effect.” Useful for medical testing problems.

Given a two-way table:

• Marginal probability: probability based on a single variable (row or column total / grand total)
• Joint probability: probability of both events occurring (cell / grand total)
• Conditional probability: probability given a condition (cell / row total or cell / column total)

A random variable assigns a numerical value to each outcome in the sample space.

A table, graph, or formula that gives the probability ($p_i$) for each value ($x_i$) of the random variable $X$.

Requirements:

1. $\sum p_i = 1$
2. $0 \leq p_i \leq 1$ for each $i$

$\mu_X = E(X) = \sum x_i \cdot p_i$

The mean of a random variable is the long-run average of its values over many repetitions.

$\sigma_X^2 = \sum (x_i - \mu_X)^2 \cdot p_i = \sum x_i^2 \cdot p_i - \mu_X^2$

$\sigma_X = \sqrt{\sigma_X^2}$

For random variables $X$ and $Y$, and constants $a$ and $b$:

• $\mu_{a + bX} = a + b\mu_X$
• $\mu_{X + Y} = \mu_X + \mu_Y$ (always)
• $\sigma_{a + bX}^2 = b^2\sigma_X^2$
• If $X$ and $Y$ are independent: $\sigma_{X + Y}^2 = \sigma_X^2 + \sigma_Y^2$ and $\sigma_{X - Y}^2 = \sigma_X^2 + \sigma_Y^2$

A binomial setting has four conditions:

1. Binary: Each observation has two possible outcomes (success/failure)
2. Independent: Observations are independent (or approximately independent if sampling without replacement and the population is at least 10 times the sample)
3. n trials: A fixed number of trials $n$
4. p probability: The probability of success $p$ is the same for each trial

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

$\mu = np, \quad \sigma = \sqrt{np(1-p)}$

A geometric setting has three conditions:

1. Binary: Each trial has two outcomes
2. Independent: Trials are independent
3. p probability: The probability of success is the same for each trial

The random variable $X$ counts the number of trials until the first success.

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

$\mu = \frac{1}{p}$

For a sufficiently large sample size $n$ (typically $n \geq 30$), the sampling distribution of $\bar{x}$ is approximately normal, regardless of the shape of the population distribution.

$\mu_{\bar{x}} = \mu, \quad \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

The CLT allows us to use normal probability calculations for sample means even when the population is not normally distributed.

For a sample proportion $\hat{p}$ with population proportion $p$ and sample size $n$:

$\mu_{\hat{p}} = p, \quad \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$

Approximately normal when $np \geq 10$ and $n(1-p) \geq 10$.

• Confusing $P(A | B)$ with $P(B | A)$
• Assuming events are independent without justification
• Confusing disjoint with independent
• Forgetting to check the 10% condition for the binomial approximation
• Misapplying the central limit theorem with small sample sizes $
• Confusing the mean of a random variable with its most probable value