In Algebra 1 you read two-way tables with row and column percents. Those percents are conditional probabilities: the row or column you restrict to is the condition.
Idea clave
In a two-way table, divide the cell by the total of the condition:
$$P(\text{row} \mid \text{column}) = \dfrac{\text{cell}}{\text{column total}}, \qquad P(\text{column} \mid \text{row}) = \dfrac{\text{cell}}{\text{row total}}.$$
With no condition, divide by the grand total.
Por qué funciona
A condition picks out one row or one column of the table: those are the only individuals we are now choosing from. The cell where the row and column meet holds the ones that satisfy both. So the denominator is always the total of whatever you were "given."
| Took review class | Did not | Total | |
|---|---|---|---|
| Passed | 45 | 90 | 135 |
| Did not pass | 5 | 60 | 65 |
| Total | 50 | 150 | 200 |
Ejemplo resuelto: Given the column
Find the probability that a student passed, given that the student took the review class.
- The condition is "took the review class": use that column, 50 students.
- Of those, 45 passed.
- $P(\text{passed} \mid \text{review}) = \dfrac{45}{50} = 0.9$.
Respuesta: $0.9$
Ejemplo resuelto: Given the row
Find the probability that a student took the review class, given that the student passed.
- The condition is "passed": use that row, 135 students.
- Of those, 45 took the class.
- $P(\text{review} \mid \text{passed}) = \dfrac{45}{135} = \dfrac{1}{3}$.
Respuesta: $\frac{1}{3}$
Ejemplo resuelto: No condition
Find the probability that a randomly chosen student took the review class and passed.
- No "given," so all 200 students are in play.
- $P = \dfrac{45}{200} = 0.225$.
Respuesta: $0.225$
Error común
Dividing by the grand total when the question gives a condition. "Of the students who took the review class, what fraction passed?" has 50 in the denominator, not 200. Check: find the word "given" (or "of those who," "among") and divide by that group's total.
Comparing $P(\text{passed} \mid \text{review}) = 0.9$ with $$P(\text{passed} \mid \text{no review}) = 90/150 = 0.6$$ shows an association between the class and passing. (This is observational data, so it does not prove the class causes passing.)