Multiple choice

What is the probability that the first $2$ draws from a pack of cards are clubs and the third is a spade? (Assume all cards are drawn without replacement)

  1. $\cfrac { 2,028 }{ 132,600 } $
  2. $\cfrac { 38 }{ 132,600 } $
  3. $\cfrac { 1,716}{ 132,600 } $
  4. $\cfrac { 2,028 }{ 140,608 } $
  5. $\cfrac { 1,716 }{ 140,608 } $
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A Correct answer
Explanation

There are 52 cards. The probability of drawing a club is 13/52. Without replacement, the probability of the second being a club is 12/51. The probability of the third being a spade is 13/50. Multiplying these: (13/52) * (12/51) * (13/50) = (1/4) * (4/17) * (13/50) = 13 / (17 * 50) = 13 / 850. Converting to the denominator 132600: (13 * 156) / (850 * 156) = 2028 / 132600.

AI explanation

Using the multiplication rule of probability for dependent events, the required probability is the product of the probabilities of drawing a club, then a second club, and then a spade. The calculations are (13/52) multiplied by (12/51) multiplied by (13/50), which equals 1/4 multiplied by 12/51 multiplied by 13/50. Multiplying the numerators gives 156, and multiplying the denominators gives 10200, so the probability is 156/10200. Multiplying both the numerator and the denominator by 13 simplifies this fraction to 2028/132600. The result is 2028/132600.