Dependence (Statistical) Math Example 3

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

Example 3

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

Solution

  1. 1
    P(1st ace)=452=113P(\text{1st ace}) = \frac{4}{52} = \frac{1}{13}
  2. 2
    P(2nd ace∣1st ace)=351=117P(\text{2nd ace}|\text{1st ace}) = \frac{3}{51} = \frac{1}{17} (3 aces remain in 51 cards)
  3. 3
    P(both aces)=113×117=1221P(\text{both aces}) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}

Answer

P(two aces)=1221≈0.0045P(\text{two aces}) = \frac{1}{221} \approx 0.0045
Drawing without replacement makes events dependent. After removing one ace, both the count of aces and total cards decrease, changing the probability for the second draw. Always adjust for cards/items already removed.

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) →

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