FP6, FP8, HP6, HP8 - Combined events and tree diagrams

Independent events

A counter is drawn, put back and mixed in before another draw. What changes if it is kept out instead? A tree makes each stage's available outcomes visible.

For independent events $A$ and $B$:

Why multiply? If $A$ occurs in a fraction $p$ of trials, and the unchanged chance of $B$ among those trials is $q$, the fraction satisfying both is $p q$.

At each tree split, the outgoing probabilities describe all possible results at that stage and sum to $1$. Multiply along a complete path to find its probability.

The diagram uses $A'$ and $B'$ for the complementary events. The two $B$-stage branch pairs agree because $A$ does not alter $B$'s chances.

[DIAGRAM: asset_name: Independent probability branches; asset_slug: independent-probability-tree; caption: The second event has probability one quarter after either first-stage result.; description: A two-stage tree has A probability three fifths and not A two fifths. After either first event, B has probability one quarter and not B three quarters. The four complete-path probabilities are three twentieths, nine twentieths, one tenth and three tenths.]

Now make the second event's probability the unknown.

Drawing from a bag with replacement and remixing, using the same uniform selection procedure each time, restores the composition and gives unchanged probabilities. With $3$ red and $2$ blue counters, two reds have probability $\frac35\times\frac35=\frac9{25}$.

Mutually exclusive is different from independent. On one die roll, odd and even are mutually exclusive; if odd occurs, even becomes impossible, so these positive-probability events are not independent.

Use different independent chances.

Dependent events

When an earlier result changes the remaining possibilities, the later probability depends on the path taken. Update the available objects before writing the next fraction.

Take a bag of $4$ red and $2$ blue counters. After a red is removed, $3$ red and $2$ blue remain. After a blue is removed, $4$ red and $1$ blue remain. Both second-stage totals are $5$, but their favourable counts differ.

[DIAGRAM: asset_name: Dependent draws without replacement; asset_slug: without-replacement-tree; caption: The first colour determines which second-stage count decreases.; description: A bag starts with four red and two blue counters. First probabilities are four sixths and two sixths. After red, second probabilities are three fifths red and two fifths blue. After blue, they are four fifths red and one fifth blue. Path probabilities are two fifths, four fifteenths, four fifteenths and one fifteenth.]