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What is the probability of drawing exactly 2 aces in a 5 card hand?

This is a hypergeometric probability question because cards are drawn without replacement from a fixed deck, so each draw changes what remains in the deck.

Probability result

3.993%

Exact fraction: 2,162/54,145

Decimal: 0.03993

The probability of exactly 2 successes is 3.993%.

You need 2 of the 4 aces and 3 of the 48 non-aces. The probability is the number of those favorable 5-card hands divided by all possible 5-card hands.

Worked steps

Show your work

  1. Inputs: population = 52, success states = 4, draws = 5, target successes = 2, event = exactly.
  2. Formula: P(X = k) = [C(K, k) C(N-K, n-k)] / C(N, n).
  3. Substitute: use C(4, k), C(48, 5-k), and C(52, 5).
  4. Steps: count the favorable hands for the requested success range, then divide by all 5-draw samples.
  5. Result: The probability of exactly 2 successes is 3.993%.

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You need 2 of the 4 aces and 3 of the 48 non-aces. The probability is the number of those favorable 5-card hands divided by all possible 5-card hands.

Formula

Probability model

P(E) = favorable outcomes / total outcomes

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Probability

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Calculation work