NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Probability experiments and simulations

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In a probability simulation conducted in Brisbane, a game involves drawing two cards consecutively without replacement from a standard 52-card deck. The game is designed so that a player wins if both cards are of the same suit.

Using a digital simulation tool, 1000 trials are conducted. What is the exact probability of winning the game in a single trial?

Express your answer as a simplified fraction.

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Worked solution and answer (click to reveal)

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Worked solution

  1. The game involves drawing two cards consecutively without replacement from a standard 52-card deck. A player wins if both cards are of the same suit. We are to find the exact probability of winning in a single trial.
  2. A standard deck has 52 cards, with 13 cards in each of the four suits: hearts, diamonds, clubs, and spades.
  3. The total number of ways to draw two cards from 52 is given by the combination formula C(n, k) = n! / (k! * (n - k)!), where n = 52 and k = 2.
  4. Calculate total outcomes: C(52, 2) = 52 * 51 / 2 = 1326.
  5. Now calculate the number of favourable outcomes: both cards from the same suit. For each suit, there are 13 cards, so C(13, 2) = 13 * 12 / 2 = 78 ways to choose two cards from one suit.
  6. There are 4 suits, so total favourable outcomes = 4 * 78 = 312.
  7. The probability of winning is the ratio of favourable outcomes to total outcomes: 312 / 1326.
  8. Simplify the fraction 312 / 1326. First, divide numerator and denominator by 6: 312 ÷ 6 = 52, 1326 ÷ 6 = 221.
  9. Now simplify 52 / 221. Check for common factors. 52 = 2 * 2 * 13, 221 = 13 * 17. So GCF is 13.
  10. Divide numerator and denominator by 13: 52 ÷ 13 = 4, 221 ÷ 13 = 17.
  11. The simplified fraction is 4 / 17.
  12. Therefore, the exact probability of winning the game in a single trial is 4/17.

Reference answer: 4/17

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