Dependence (Statistical) Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Dependence (Statistical).

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

When the probability of one event changes based on whether another event occurred.

Knowing A happened tells you something about B—they're connected.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Dependence requires conditional probability; independence allows multiplication.

Common stuck point: Dependence \neq causation. Rain and umbrellas are dependent but rain doesn't cause umbrellas.

Sense of Study hint: Compare P(B) with P(B|A). If they differ, the events are dependent. Use the multiplication rule P(A) * P(B|A) for the joint probability.

Worked Examples

Example 1

medium
A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find P(\text{both red}) using the multiplication rule for dependent events.

Solution

  1. 1
    Event A = first ball is red: P(A) = \frac{5}{8}
  2. 2
    Event B = second ball is red, given first was red: P(B|A) = \frac{4}{7} (only 4 red left among 7)
  3. 3
    Apply multiplication rule: P(A \cap B) = P(A) \cdot P(B|A) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}
  4. 4
    Note: events are dependent because removing the first ball changes the composition of the bag

Answer

P(\text{both red}) = \frac{5}{14} \approx 0.357
Without replacement creates dependence — the outcome of the first draw changes the probabilities for the second. The general multiplication rule P(A \cap B) = P(A) \cdot P(B|A) handles both dependent and independent events.

Example 2

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

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
A deck of 52 cards. Find P(\text{drawing two aces in a row}) without replacement.

Example 2

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

These ideas may be useful before you work through the harder examples.

probabilityindependent events