Independent events

Events $A$ and $B$ are independent when $P(A \cap B) = P(A)P(B)$. Provided $P(B) > 0$, that is the same as $P(A \mid B) = P(A)$, read "the probability of A given B equals the probability of A", and the relation runs both ways: if $B$ tells you nothing about $A$, then $A$ tells you nothing about $B$. Independence is a numerical condition you check, not a verdict you reach from how the story sounds.

It does not require two separate physical actions. Roll one fair die, and let $A$ be "even", the set $\{2, 4, 6\}$, and $B$ be "at least 5", the set $\{5, 6\}$. Then $P(A) = 0.5$, $P(B) = 2/6 \approx 0.3333$, and $A \cap B$ is the single outcome 6, so $P(A \cap B) = 1/6 \approx 0.1667$. The product $0.5 \times 1/3 = 1/6$ matches it, so two events on the very same roll are independent, and indeed $P(A \mid B) = (1/6)/(2/6) = 0.5 = P(A)$. Widen $B$ to "at least 4" and it breaks: $P(A \cap B) = 2/6 \approx 0.3333$ against a product of $0.5 \times 0.5 = 0.25$.

"The two events cannot both happen, so one cannot affect the other, so they are independent." That gets it exactly backwards. Mutually exclusive events with nonzero probabilities are never independent, because $P(A \cap B) = 0$ while $P(A)P(B) > 0$, and those two cannot be equal. If $A$ and $B$ are disjoint, $B$ occurring forces $A$ not to occur, which is the strongest dependence available.

Selecting people without replacement destroys independence. In a class of 20 with 8 in band, picking two students at random gives $P(\text{second in band} \mid \text{first in band}) = 7/19 \approx 0.3684$, not the $8/20 = 0.40$ you started with, because the first pick shrinks the class. The ten percent condition is what lets such draws be treated as close enough to independent. Independence itself is topic 2.7.