Probability & Statistics

If a task is made of several independent stages, and stage $1$ can happen in $m$ ways, stage $2$ in $n$ ways, and so on, then the whole task can happen in

This is the engine behind every counting formula in this topic: whenever choices are made in sequence, you multiply.

Both count ways to choose $r$ items from $n$ distinct items.

• Permutation — order matters (arrangements, rankings, seating): @@BLOCK0@@
• Combination — order does not matter (committees, teams, hands): @@BLOCK1@@

Because a combination ignores the $r!$ orderings of each group, $\binom{n}{r} = \dfrac{P(n,r)}{r!}$.

For equally likely outcomes,

Theoretical probability comes from the model (a fair die gives $\tfrac16$). Experimental probability comes from data: $\dfrac{\text{times it happened}}{\text{trials}}$. The complement rule saves work:

For the probability that both $A$ and $B$ happen:

• Independent — one outcome doesn't affect the other, so $P(B \mid A) = P(B)$ and the rule becomes $P(A)\cdot P(B)$. (Coin flips, drawing with replacement.)
• Dependent — the first outcome changes the second (drawing without replacement).

Conditional probability isolates $P(B \mid A) = \dfrac{P(A \text{ and } B)}{P(A)}$.

For the probability that $A$ or $B$ happens:

• Mutually exclusive (disjoint) — can't both happen, so $P(A \text{ and } B) = 0$ and the rule is just $P(A)+P(B)$.
• Overlapping — can happen together, so subtract the overlap once (it was counted twice).

A binomial experiment has: (1) a fixed number of trials $n$, (2) two outcomes per trial — success or failure, (3) a constant success probability $p$, and (4) independent trials. The probability of exactly $k$ successes is

Here $\binom{n}{k}$ counts which trials succeed, $p^k$ is their successes, and $(1-p)^{n-k}$ the remaining failures.

Center: the mean $\bar{x} = \dfrac{\sum x}{n}$, the median (middle value when ordered), and the mode (most frequent). Spread: the variance $\sigma^2$ is the average squared distance from the mean, and the standard deviation $\sigma$ is its square root:

Standard deviation is in the same units as the data, so it measures “typical distance from the mean.”

Many natural measurements (heights, test scores, errors) follow a normal distribution: a symmetric, bell-shaped curve centered at the mean $\mu$, with spread set by $\sigma$. For any normal distribution, the Empirical Rule (68–95–99.7) says:

• about $68\%$ of data lies within $1\sigma$ of the mean,
• about $95\%$ within $2\sigma$,
• about $99.7\%$ within $3\sigma$.

A z-score rewrites any value $x$ as its distance from the mean, measured in standard deviations:

A positive $z$ is above the mean, negative is below, and $z = 0$ is exactly average. z-scores let you compare values from different distributions on one common scale.

Not every arrangement is a plain permutation — watch for two twists.

• Circular arrangements. Seating $n$ people around a round table has no “first” seat, so every arrangement can be rotated $n$ ways to look the same. Divide the overcount away: @@BLOCK0@@
• Identical (repeated) items. Arranging $n$ objects where one kind repeats $n_1$ times, another $n_2$ times, and so on: @@BLOCK1@@ We divide by each $n_i!$ because swapping identical items produces no new word.

When a direct count splits into many cases, count the opposite and subtract from the total. For probabilities this is the complement rule; for raw counts it is the same idea:

The phrase “at least one” is the signal: the negation “none” is almost always a single, clean computation.

Conditional probabilities are not symmetric: $P(A \mid B) \ne P(B \mid A)$ in general. Bayes' theorem flips a known conditional into the one you want:

The denominator is the law of total probability: it rebuilds $P(B)$ by splitting across $A$ and its complement $A^c$.

The expected value of a numerical outcome $X$ weights each value by its probability:

It is the average you would approach over many repetitions — not necessarily a possible single outcome. A game is fair when $E(\text{net gain}) = 0$.

For a binomial with $n$ trials and success probability $p$, a threshold question sums several exact-$k$ terms:

When $k$ is small, the complement is faster: $P(X \ge 1) = 1 - P(X = 0)$, and $P(X \ge 2) = 1 - P(X=0) - P(X=1)$.

Shifting and scaling every data value changes center and spread in predictable ways. If $y = ax + b$:

• Mean shifts and scales: $\bar{y} = a\bar{x} + b$.
• Standard deviation scales but does not shift: $\sigma_y = |a|\,\sigma_x$.
• Variance scales by the square: $\sigma_y^2 = a^2\,\sigma_x^2$.

Adding a constant $b$ slides all data together, so spread is unchanged; multiplying by $a$ stretches the gaps by $|a|$.

To pool two groups of sizes $n_1, n_2$, the combined mean is the weighted average — never the plain average of the two means unless the groups are the same size:

Variance must be pooled about the combined mean. Using each group's sum of squared deviations $S_i = n_i \sigma_i^2$ and combined mean $\bar{x}_c$:

The extra $(\bar{x}_i - \bar{x}_c)^2$ terms account for how far each group's center sits from the shared center.