Dependence (Statistical) Math Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

hard
Disease test: P(disease)=0.05P(\text{disease}) = 0.05. Test positive given disease: P(+โˆฃD)=0.90P(+|D) = 0.90. Test positive given no disease: P(+โˆฃDc)=0.10P(+|D^c) = 0.10. Find P(Dโˆฉ+)P(D \cap +) and P(Dcโˆฉ+)P(D^c \cap +).

Solution

  1. 1
    P(Dโˆฉ+)=P(D)โ‹…P(+โˆฃD)=0.05ร—0.90=0.045P(D \cap +) = P(D) \cdot P(+|D) = 0.05 \times 0.90 = 0.045
  2. 2
    P(Dcโˆฉ+)=P(Dc)โ‹…P(+โˆฃDc)=0.95ร—0.10=0.095P(D^c \cap +) = P(D^c) \cdot P(+|D^c) = 0.95 \times 0.10 = 0.095
  3. 3
    Total positive tests: P(+)=0.045+0.095=0.14P(+) = 0.045 + 0.095 = 0.14
  4. 4
    Interpretation: Of those testing positive, only 0.0450.14โ‰ˆ32%\frac{0.045}{0.14} \approx 32\% actually have the disease!

Answer

P(Dโˆฉ+)=0.045P(D \cap +) = 0.045; P(Dcโˆฉ+)=0.095P(D^c \cap +) = 0.095. Most positives are false positives.
Dependent events require conditional probabilities. In medical testing, test results depend on disease status. The counter-intuitive result (most positives are false positives) arises because disease is rare โ€” even a good test produces many false positives when base rate is low.

About Dependence (Statistical)

Two events are statistically dependent when knowing one event occurred changes the probability of the other โ€” formally, P(BโˆฃA)โ‰ P(B)P(B|A) \neq P(B), meaning the events share information.

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More Dependence (Statistical) Examples

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