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Learn through experiments

Check whether conditioning changes a probability

Are independence and mutual exclusion the same?

Entering midway? A joint event means both A and B occur. A conditional probability restricts attention to the cases where another event occurred.

1 · Try

Start with a concrete question

An original synthetic population table controls A prevalence and B rates within and outside A. Four cells partition the whole population.

Try the initial inputs, then write a prediction before changing one setting. All calculations run locally on your CPU.

Go directly to the experiment · Review prerequisite mathematics and return here

2 · Predict

What do you expect?

Predict what equality must hold for A and B to be independent.

3 · Experiment

Change one thing. Inspect what happens.

  1. Calculate all four cell probabilities and confirm they sum to one.
  2. Make B rates equal inside and outside A.
  3. Set B inside A to zero but B outside A positive; explain why mutual exclusion is dependence.
Experiment inputs
Inputs
Your settingsMechanism
Calculation belowEvidence
Compare outcomes
P(B)50 %
P(B given A)80 %
P(A and B)40 %
Independent-product comparison25 %

Compare the current results

Every bar starts at zero and uses the same numeric scale in this chart. The upper bound automatically adjusts to the largest current value (at least 1), so compare the numbers when changing inputs. Units and quantities come from each label; different kinds of quantity are not interchangeable. Displayed values are rounded to at most six significant digits.
040
  • A and B (%)40
    A and B (%): 40Horizontal bar from zero to 40, on the shared zero to 40 numeric axis.
  • A and not B (%)10
    A and not B (%): 10Horizontal bar from zero to 10, on the shared zero to 40 numeric axis.
  • Not A and B (%)10
    Not A and B (%): 10Horizontal bar from zero to 10, on the shared zero to 40 numeric axis.
  • Neither (%)40
    Neither (%): 40Horizontal bar from zero to 40, on the shared zero to 40 numeric axis.

Follow the calculation

  1. A fraction 0.5; B within A 0.8; B outside A 0.2.
  2. P(B) = 0.5 × 0.8 + 0.5 × 0.2 = 0.5.
  3. P(A and B) = 0.5 × 0.8 = 0.4.
  4. If independent, joint would equal P(A)P(B) = 0.25.

Independence means conditioning on A does not change the probability of B. Mutually exclusive events with positive probability are not independent.

Captures keep your inputs, prediction and observation. Results are recalculated from those inputs. Resetting inputs keeps your captures and challenge history.

4 · Understand

Explain the mechanism

Check whether conditioning changes a probability

Recap for entering here

A joint event means both A and B occur. A conditional probability restricts attention to the cases where another event occurred.

The concrete question

An original synthetic population table controls A prevalence and B rates within and outside A. Four cells partition the whole population.

Predict what equality must hold for A and B to be independent.

Work through the mechanism

At defaults P(A)=0.5, P(B|A)=0.8 and P(B|not A)=0.2. P(B)=0.5, while P(A and B)=0.4.

Independence would require P(A)P(B)=0.25 to equal the joint 0.4; it does not.

If positive-probability events cannot occur together, learning that A occurred makes B impossible, so they are dependent.

Formula and symbols

P(B)=P(A)P(B|A)+P(not A)P(B|not A). Independence requires P(A and B)=P(A)P(B).

Test the idea

  1. Calculate all four cell probabilities and confirm they sum to one.

  2. Make B rates equal inside and outside A.

  3. Set B inside A to zero but B outside A positive; explain why mutual exclusion is dependence.

Assumptions and boundaries

  • This is an authored probability table, not a significance test of empirical independence.

  • No causal relationship follows from dependence alone.

Extend your investigation

High school: Compare two-die events with mutually exclusive faces on one die.

College: Test independence by comparing the joint distribution with the product of marginals.

Working professionals: List dependencies that would invalidate multiplying component reliability probabilities.

5 · Apply

Try a fresh case

Independent events have probabilities 0.3 and 0.4. What is the probability both occur?

This is a public practice check. Retries support learning; matching a number does not certify independent proficiency.

Take the investigation further

Choose any extension. These are project prompts for you to carry out; they are not automatically assessed.

School investigation

Compare two-die events with mutually exclusive faces on one die.

College investigation

Test independence by comparing the joint distribution with the product of marginals.

Workplace investigation

List dependencies that would invalidate multiplying component reliability probabilities.

6 · Continue

Keep the evidence, choose your next step

Completion records your own decision that you have worked through this investigation. Practice checks and independent project evidence are separate.