In a standard deck, what is the probability of drawing two aces in succession without replacement?

Unlock all questions

This demo includes only 20 questions. Upgrade to access hundreds of questions, flashcards, exam simulations, and disable ads.

Full question bankExam simulationsFlashcards
From $24.99Unlock all

Prepare for the AMSOC 26-003 Module A Test. Utilize flashcards and multiple choice questions with hints and explanations. Ace your exam!

Multiple Choice

In a standard deck, what is the probability of drawing two aces in succession without replacement?

Explanation:
When drawing two cards without replacement, the chance of both being aces is found by multiplying the probabilities of each step, because the second event depends on what happened first. First card is an ace: 4 out of 52, which is 4/52 = 1/13. If that happens, there are now 3 aces left among 51 cards, so the second card is an ace with probability 3/51 = 1/17. Multiply these: (1/13) × (1/17) = 1/221. Another way to see it is counting: there are 4×3 = 12 ordered ace-ace pairs possible, out of 52×51 = 2652 possible ordered two-card sequences. 12/2652 simplifies to 1/221. So the probability is 1/221.

When drawing two cards without replacement, the chance of both being aces is found by multiplying the probabilities of each step, because the second event depends on what happened first.

First card is an ace: 4 out of 52, which is 4/52 = 1/13. If that happens, there are now 3 aces left among 51 cards, so the second card is an ace with probability 3/51 = 1/17. Multiply these: (1/13) × (1/17) = 1/221.

Another way to see it is counting: there are 4×3 = 12 ordered ace-ace pairs possible, out of 52×51 = 2652 possible ordered two-card sequences. 12/2652 simplifies to 1/221.

So the probability is 1/221.

More Multiple Choice Questions
Subscribe

Get the latest from Passetra

You can unsubscribe at any time. Read our privacy policy