Dependence (Statistical) Math Example 1

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

Example 1

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

Solution

  1. 1
    Event A = first ball is red: P(A)=58P(A) = \frac{5}{8}
  2. 2
    Event B = second ball is red, given first was red: P(B∣A)=47P(B|A) = \frac{4}{7} (only 4 red left among 7)
  3. 3
    Apply multiplication rule: P(A∩B)=P(A)⋅P(B∣A)=58×47=2056=514P(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(both red)=514≈0.357P(\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∩B)=P(A)⋅P(B∣A)P(A \cap B) = P(A) \cdot P(B|A) handles both dependent and independent events.

About Dependence (Statistical)

Two events are statistically dependent when knowing one event occurred changes the probability of the other — formally, P(B∣A)≠P(B)P(B|A) \neq P(B), meaning the events share information.

Learn more about Dependence (Statistical) →

More Dependence (Statistical) Examples

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.

Example 4 hard

Verify whether smoking and lung cancer are dependent using the following: [formula], [formula]. What

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