Dependence (Statistical) Math Example 4

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

Example 4

hard
Verify whether smoking and lung cancer are dependent using the following: P(cancer)=0.06P(\text{cancer}) = 0.06, P(cancerโˆฃsmoker)=0.15P(\text{cancer}|\text{smoker}) = 0.15. What does this tell us about the relationship?

Solution

  1. 1
    For independence, we need P(cancerโˆฃsmoker)=P(cancer)P(\text{cancer}|\text{smoker}) = P(\text{cancer})
  2. 2
    Check: P(cancerโˆฃsmoker)=0.15โ‰ 0.06=P(cancer)P(\text{cancer}|\text{smoker}) = 0.15 \neq 0.06 = P(\text{cancer})
  3. 3
    Conclusion: smoking and cancer are dependent โ€” knowing someone smokes increases the probability of cancer
  4. 4
    Note: dependence shows association, but correlation โ‰  causation; further evidence needed for causal claim

Answer

Dependent: P(cancerโˆฃsmoker)=0.15โ‰ 0.06P(\text{cancer}|\text{smoker}) = 0.15 \neq 0.06. Smoking is associated with higher cancer risk.
Dependence is detected when conditional probability differs from marginal probability. Statistical dependence means the events are associated, but establishing causation requires experimental evidence (not just observational data).

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.

Learn more about Dependence (Statistical) โ†’

More Dependence (Statistical) Examples

Example 1 medium

A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find [formula] using the

Example 2 hard

Disease test: [formula]. Test positive given disease: [formula]. Test positive given no disease: [fo

Example 3 easy

A deck of 52 cards. Find [formula] without replacement.

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