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Independent Events and Unions of Events

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Events are independent when conditioning changes nothing, tested equivalently by $P(A \mid B) = P(A)$, $P(B \mid A) = P(B)$, or $P(A \text{ and } B) = P(A)P(B)$. The general multiplication rule $P(A \text{ and } B) = P(A)P(B \mid A)$ always holds, and it shortens to a plain product exactly when independence has been verified. Drawing with replacement makes successive draws independent; drawing without replacement makes them dependent, though the 10% condition says the approximation is fine when the sample is at most a tenth of the population. Unions still take the general addition rule, with independence supplying the intersection when it is not given.

The failures are the shortcuts. Unconditional probabilities are multiplied for dependent events, most often draws without replacement, where $\frac{4}{10} \times \frac{4}{10}$ replaces $\frac{4}{10} \times \frac{3}{9}$. Independence is assumed because two events sound unrelated, rather than checked against one of the three tests. Mutually exclusive is used as an independence test, though disjointness with positive probability guarantees dependence. And at least one is computed by adding, which double counts the overlaps and can exceed 1, where $1 - P(\text{none})$ handles any number of independent events in one step.

400 adults smart TV no TV total subscriber 132 108 240 not 88 72 160 total 220 180 400 P(TV | subscriber) = 132/240 = 0.55 P(TV) = 220/400 = 0.55, so INDEPENDENT third test agrees: 0.60 x 0.55 = 0.33 = 132/400
Independence is checked, not felt. Conditioning on subscriber leaves the smart-TV rate at 0.55, and the product test lands on the same joint proportion the table reports.
with replacement: independent 4/10 6/10 4/10 6/10 RR P(both red) = 0.40 x 0.40 = 0.16 the bag is restored, so the second draw forgets the first without replacement: dependent 4/10 6/10 3/9 6/9 RR P(both red) = 0.40 x 3/9 = 0.133 one red is gone and the bag is smaller: 3 of 9 remain
The only difference between the trees is the second branch. Sampling without replacement makes it conditional, and using the unconditional value there is the standard wrong answer.

The work

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Lesson
Independent Events and Unions of Events ›

Gives independence three equivalent tests and insists one of them be run, separates the general multiplication rule from the independent shortcut, works with and without replacement, and routes at least one through the complement.

Skill check · 10 scenarios
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Diagnostic
10-item topic check ›

Ten items on independence and unions: unconditional probabilities multiplied for dependent draws, independence assumed from the story, disjoint used as an independence test, and at least one computed by adding. Take it cold to find your habit, or after the lesson to check it is gone.

Not started · 10 items · ~15 min
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Targeted Practice
Drill a single misconception ›

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

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