Multiple choice

Cards are drawn one by one without replacement from a pack of $52$ cards. The probability that $10$ cards will precede the first ace is

  1. $241/1456$
  2. $164/4165$
  3. $451/884$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The probability that the first ace appears at position 11 is the number of ways to arrange 48 non-aces in the first 10 spots and an ace in the 11th, divided by total permutations. This is C(48, 10) / C(52, 10) * (4/42) = 164/4165.

AI explanation

For 10 non-aces to precede the first ace, the first 11 cards drawn must consist of 10 non-aces followed by 1 ace. The total number of ways to choose the first 11 positions from the 48 non-aces and 4 aces is found by combinations: 48C10 times 38C1 times 41 (to arrange the remaining cards), divided by the total ways to arrange the deck, 52C11 times 41. This simplifies to the ratio of ways to pick 10 non-aces from 48 and 1 ace from 4, which is 48C10 times 4, divided by the ways to pick any 11 cards from 52, which is 52C11. Calculating (48! divided by (38!10!)) times 4 divided by (52! divided by (41!11!)) yields 164/4165.

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