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.

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• 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:

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• 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:

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• 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.”

One outlier drags the mean but not the median.

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:

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

One outlier drags the mean but not the median.

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.

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• 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.

One outlier drags the mean but not the median.

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 $y = ax + b$:

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• 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.