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What is the probability of drawing exactly 1 ace 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

29.9474%

Exact fraction: 3,243/10,829

Decimal: 0.299474

The probability of exactly 1 success is 29.9474%.

You need 1 of the 4 aces and 4 of the 48 non-aces. The probability is the number of those favourable 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 = 1, 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 1 success is 29.9474%.

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You need 1 of the 4 aces and 4 of the 48 non-aces. The probability is the number of those favourable 5-card hands divided by all possible 5-card hands.

Formula

Probability model

P(E) = favorable outcomes / total outcomes

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Calculation work