Probability Topics

3.1 Chance and long-run frequency

Flip a coin once and you cannot say what it will show. Flip it a thousand times and you can say, quite precisely, what fraction will be heads. Probability is that long-run fraction. Suppose a coin lands heads with probability \(p\). After \(n\) flips the relative frequency of heads, \(\hat{p} = \dfrac{\text{heads}}{n}\), settles toward \(p\) as \(n\) grows. This is the law of large numbers.

Set \(p\) with the slider and flip. The running proportion swings widely in the first ten or twenty flips, then calms down and hugs the dashed line. It never lands exactly on \(p\) and stays there. It only wanders less and less. A short run of heads is not evidence of a biased coin. A long run of many flips is.

Reading: OpenStax Introductory Statistics, Section 3.1.

3.2 Events, unions, intersections

The rectangle is the sample space. Every dot is one possible outcome, and each dot is equally likely. An event is a region of it, like circle A. The intersection “A and B” is where the circles overlap. The union “A or B” is everything inside either circle. The complement of A is everything outside it. Counting dots gives \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), because dots in the overlap would otherwise be counted twice.

Drag the circles apart until no dot is in both. Now A and B are mutually exclusive, \(P(A \cap B) = 0\), and the union is just the sum. That is not the same as independent, which means \(P(A \cap B) = P(A)\,P(B)\). Compare the two readouts as you move the circles. Overlapping circles can be close to independent, while separated ones never are.

Reading: OpenStax Introductory Statistics, Sections 3.2, 3.3, 3.5.

3.3 Conditional probability and contingency tables

Each ball is a student. First it splits by whether the student studied, then by whether they passed. The conditional probability \(P(\text{passed} \mid \text{studied})\) asks a narrower question. Among the students who studied, what share passed? It is computed from the counts as

\[P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{n(A \text{ and } B)}{n(B)}.\]

Press “Show only studied”. The other branch fades, and the sample space shrinks to the studied row of the contingency table. The conditional probability is just an ordinary probability inside that smaller world. Slide the two “passed” probabilities until they match. Studying then tells you nothing about passing, and the events are independent.

Reading: OpenStax Introductory Statistics, Sections 3.4, 3.5.