Independence is unchanged chance
Knowing one event occurred does not change the probability of the other. Use exact probabilities to test the model. A diagram alone cannot decide independence
- Product test Compare the probability of both events with the product of their separate probabilities. Equality means independent; inequality means dependent \(\displaystyle \Pr(A\cap B)=\Pr(A)\Pr(B)\)
- Conditioning test When B has positive probability, compare A within B with A in the whole sample space \(\displaystyle \Pr(A\mid B)=\Pr(A)\)
- Exclusive is different Two positive-probability events that cannot occur together are dependent. Knowing one occurred rules out the other
- State the model Counts give exact model probabilities only when individual outcomes are equally likely. Observed frequencies instead give estimates
Worked example
One of 30 equally likely cards is selected. A contains 12 cards, B contains 10 and the overlap contains 4. Test independence
- \(\displaystyle \Pr(A)=12/30=2/5\)Include the whole of A.
- \(\displaystyle \Pr(B)=10/30=1/3\)Include the whole of B.
- \(\displaystyle \Pr(A\cap B)=4/30=2/15\)Use the overlap once
- \(\displaystyle \begin{aligned}&\Pr(A)\Pr(B)\\&=(2/5)(1/3)\\&=2/15\end{aligned}\)The product equals the overlap probability, so the events are independent
- \(\displaystyle \Pr(A\mid B)=4/10=2/5\)The conditional test agrees: knowing B leaves the chance of A unchanged
Common mistakes
Difficulty levels
Questions about this topic
Do overlapping circles show dependence?+
No. Venn circles show membership; exact counts or probabilities decide independence. Their drawn areas need not be proportional
Can independent events happen together?+
Yes. Their overlap probability equals the product of their marginal probabilities
Are mutually exclusive events independent?+
Not if both have positive probability. An event with probability zero is independent of every event under the product test, even when they are disjoint
Why compare A given B with A?+
The comparison asks whether learning B changes the chance of A.
Can I reverse A and B?+
Yes for independence. Conditional probabilities themselves usually differ; each uses its own conditioning denominator
What if B has probability zero?+
A given B is undefined in this elementary ratio model. Do not divide by zero; the product test remains meaningful
Does equality of rounded decimals prove independence?+
No. Use exact fractions; different values can round to the same decimal
Do similar experimental frequencies prove independence?+
No. They can support a model, but finite runs fluctuate and do not establish exact equality
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 10 Mathematics, chapter 4: Probability