LESSON 1 · 19 MIN
Probability and statistics
Model the complete outcome space, preserve conditional and sequential denominators, and compute center or spread from complete, correctly weighted data.
ESSENTIAL QUESTIONWhat is the active denominator—and which summary answers the comparison without hiding the underlying data?
Restrict the sample space before counting what is favorable.
| Color | Circles | Squares | Row total |
|---|---|---|---|
| Red | 3 | 2 | 5 |
| Blue | 1 | 6 | 7 |
| Column total | 4 | 8 | 12 |
| Audit step | Included token classes | Exact count | Role |
|---|---|---|---|
| Original space | All four color-shape classes | 12 | Superseded denominator after conditioning |
| Given square | Red squares and blue squares | 8 | Active denominator |
| Red and square | Red squares only | 2 | Favorable numerator |
| Path step | Blue tokens available | Total tokens available | Exact factor |
|---|---|---|---|
| First draw is blue | 3 | 8 | 3/8 |
| Second draw is blue | 2 | 7 | 2/7 |
| Two blue draws | Dependent factors multiplied | 3/28 | |
“Without replacement” changes the second branch from 3/8 to 2/7. The result is available entirely through the tables and does not depend on color.
Recover totals, then compare center and spread separately.
| Group | Count | Group mean | Recovered group total |
|---|---|---|---|
| Section A | 8 | 72 | 576 |
| Section B | 12 | 84 | 1008 |
| Combined | 20 | 396/5 = 79.2 | 1584 |
| Ordered position | Distribution A | Distribution B |
|---|---|---|
| 1 | 2 | 4 |
| 2 | 4 | 5 |
| 3 | 6 | 6 |
| 4 | 8 | 7 |
| 5 | 10 | 8 |
| Summary | Distribution A | Distribution B | Comparison |
|---|---|---|---|
| Mean | 6 | 6 | Equal centers |
| Median | 6 | 6 | Equal centers |
| Range | 8 | 4 | Distribution A is more spread by range |
Equal means and medians do not make two distributions identical. The complete values show that A extends farther from the shared center.
Condition before dividing
A condition reduces the sample space first; the event count is then measured inside that reduced denominator.
- Write the complete outcome inventory before selecting favorable cases.
- P(A given B) uses B—not the original sample space—as its denominator.
Update dependent stages
Without replacement, both the category count and total count change after a draw, so repeated factors are not automatically equal.
- At least one is often one minus the probability of none.
- Replacement preserves counts; no replacement changes the next branch.
Pair center with spread
A mean, median, and range describe different features, and pooled means must weight each group by its actual count.
- The mean of group means is valid only when group sizes are equal.
- Equal centers do not imply equal variability or identical data.
Worked example
A bag has 5 red and 3 blue tokens. Two are drawn without replacement. What is the probability of at least one red token?
- 1
Use the complement: at least one red equals one minus no red, which means two blue draws.
- 2
The first blue probability is 3/8. After a blue draw without replacement, the next blue probability is 2/7.
- 3
Multiply the dependent complement path: 3/8 × 2/7 = 3/28, then subtract from 1.
ConclusionThe exact probability is 25/28. Reusing 3/8 for the second draw would incorrectly assume replacement.
Calculator decision ledger
- Model mentally
- List the active outcomes or complete data rows, label the denominator, and identify dependence or group weights before entering values.
- Estimate
- Predict whether the result should exceed one-half and which data set should be more spread before exact arithmetic.
- Calculator
- Use exact fractions for short probability paths and calculator arithmetic only for longer weighted totals or summary checks.
- Audit
- Confirm that probabilities stay between zero and one, frequencies sum to the stated count, and every summary uses the intended denominator.
Close the notes first
Retrieve the denominator.
01What becomes the denominator in P(A given B)?
Conditioning removes outcomes outside B before A is evaluated.
02What changes after a draw without replacement?
The next event is dependent on which object left the population.
03How do unequal group sizes affect a combined mean?
An unweighted average gives small and large groups equal influence.