Sometimes knowing B changes the chance of A (as with the review class and passing). Sometimes it changes nothing at all. That second case has a name, and it builds directly on conditional probability.
Idea clave
Events A and B are independent if knowing one happened does not change the probability of the other: $P(A \mid B) = P(A)$. Equivalently, $P(A \cap B) = P(A) \cdot P(B)$. Otherwise they are dependent.
Por qué funciona
Start from the definition $P(A \mid B) = P(A \cap B)/P(B)$. If $P(A \mid B) = P(A)$, multiply both sides by $P(B)$ to get $P(A \cap B) = P(A)P(B)$; the steps reverse, so the two tests say the same thing. In words: A happens in the same fraction of the B-world as of the whole world.
Ejemplo resuelto: Checking a table
Of 200 students, 60 play an instrument. Among the 80 ninth graders, 24 play one. Are "plays an instrument" and "ninth grader" independent?
- $P(\text{instrument}) = \dfrac{60}{200} = 0.3$.
- $$P(\text{instrument} \mid \text{9th grade}) = \dfrac{24}{80} = 0.3$$
- Equal: knowing the grade does not change the chance.
Respuesta: Independent.
Ejemplo resuelto: Checking with the product
$P(A) = 0.6$, $P(B) = 0.5$, $P(A \cap B) = 0.2$. Are A and B independent?
- If independent, $P(A \cap B)$ would be $0.6 \times 0.5 = 0.3$.
- It is $0.2$, not $0.3$.
Respuesta: Not independent.
Ejemplo resuelto: Mutually exclusive is not independent
A = "the die shows 1" and B = "the die shows 6." Are they independent?
- $P(A) = \frac{1}{6}$ and $P(B) = \frac{1}{6}$.
- They cannot both happen: $P(A \cap B) = 0$, but $P(A)P(B) = \frac{1}{36}$.
- Knowing the die shows 1 changes the chance of 6 from $\frac{1}{6}$ to 0.
Respuesta: Dependent (and mutually exclusive).
Error común
Thinking mutually exclusive events are independent. They are the opposite: if A happens, B cannot, so knowing about A tells you everything about B. Check: if $P(A \cap B) = 0$ but both $P(A)$ and $P(B)$ are positive, the events are dependent.