Independence is a test, not a feeling

Events $A$ and $B$ are independent when knowing that one occurred does not change the probability of the other. Three statements say this, and if any one holds, all three do:

1. $P(A \mid B) = P(A)$
2. $P(B \mid A) = P(B)$
3. $P(A \text{ and } B) = P(A) \cdot P(B)$

Of 400 adults, 240 subscribe to a streaming service and 220 own a smart TV; 132 do both. Then $P(\text{TV} \mid \text{subscriber}) = \frac{132}{240} = 0.55$ and $P(\text{TV}) = \frac{220}{400} = 0.55$. Conditioning changed nothing, so the two are independent in this sample, and the third test agrees: $0.60 \times 0.55 = 0.33 = \frac{132}{400}$.

Two cautions. Independence is a numerical property, not a judgment about whether the topics are connected, so it must be checked rather than assumed from the story. And independence is not mutual exclusivity: disjoint events with positive probability are maximally dependent, because one occurring drops the other to 0.

The general multiplication rule follows from the definition of conditional probability and holds for any two events:

$$P(A \text{ and } B) = P(A) \cdot P(B \mid A).$$

When $A$ and $B$ are independent, $P(B \mid A) = P(B)$ and the rule shortens to $P(A)P(B)$. That shortcut is the single most over-applied line in the unit. Using it on dependent events makes an answer that is confidently wrong, and there is no warning in the arithmetic.

Extending to several events, independent events multiply straight through: for three independent events, $P(A \text{ and } B \text{ and } C) = P(A)P(B)P(C)$. Dependent events chain their conditions instead: $P(A)P(B \mid A)P(C \mid A \text{ and } B)$.

A bag holds 4 red and 6 blue chips. Drawing two:

1. With replacement: the first chip goes back, so the bag is identical for the second draw and $P(\text{both red}) = \frac{4}{10} \times \frac{4}{10} = 0.16$. The draws are independent.
2. Without replacement: the first chip is gone, so $P(\text{both red}) = \frac{4}{10} \times \frac{3}{9} \approx 0.133$. The draws are dependent, and the second probability is conditional on the first.

Real sampling is almost always without replacement, so draws from a population are strictly dependent. When the sample is small relative to the population, the dependence is negligible, and the standard rule of thumb is the 10% condition: if the sample is no more than 10% of the population, treating the draws as independent is a good approximation. Sampling 5 people from a town of 50,000 barely changes the composition; sampling 5 from a club of 12 changes it a great deal.

The general addition rule is unchanged: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$. What independence adds is a way to obtain the intersection when it is not given: substitute $P(A)P(B)$.

If a flight is delayed with probability 0.20 and a connecting flight is delayed with probability 0.15, independently, then

$$P(\text{at least one delay}) = 0.20 + 0.15 - (0.20)(0.15) = 0.32,$$

and the complement route gives the same value in one step: $1 - (0.80)(0.85) = 1 - 0.68 = 0.32$. For more than two independent events the complement is the only practical route: with five independent events at 0.20 each, $P(\text{at least one}) = 1 - (0.80)^5 \approx 0.672$, while adding gives $5 \times 0.20 = 1.00$, which is wrong and, one event later, impossible.

Keep the two rules attached to their words. "And" multiplies, with a conditional factor unless independence is established. "Or" adds, minus the overlap unless the events are disjoint. The two errors that follow from dropping those qualifiers, a product where a conditional belonged and a sum where an overlap belonged, are the most common wrong answers in this unit.