Two cards are drawn one after another without replacement from a standard pack of 52 playing cards. What is the probability that both cards drawn are Aces?

Introduction / Context:
This probability question involves drawing cards without replacement from a standard deck. We want the probability that both cards drawn are Aces, which requires sequential probabilities with a changing sample space after the first draw.

Given Data / Assumptions:

• A standard deck has 52 cards.
• There are 4 Aces in the deck.
• Two cards are drawn one after another without replacement.
• We are interested in the event that both drawn cards are Aces.

Concept / Approach:
We use the multiplication rule for dependent events. The probability that the first card is an Ace is 4 out of 52. After drawing one Ace, there are 3 Aces left and only 51 cards remaining, so the probability that the second card is also an Ace is 3 out of 51. The probability that both events occur is the product of these two probabilities.

Step-by-Step Solution:
Step 1: Probability that the first card drawn is an Ace = 4 / 52. Step 2: After one Ace is drawn, remaining Aces = 3 and total remaining cards = 51. Step 3: Probability that the second card drawn is an Ace given the first was an Ace = 3 / 51. Step 4: Multiply to find the joint probability for both draws. P(both Aces) = (4 / 52) * (3 / 51). Step 5: Simplify: 4 / 52 = 1 / 13, and 3 / 51 = 1 / 17, since 51 = 3 * 17. Step 6: Therefore, P(both Aces) = (1 / 13) * (1 / 17) = 1 / 221.

Verification / Alternative check:
You can treat the sample space as ordered draws from 52 cards and confirm that the favourable outcomes are sequences of two Aces. There are 4 choices for the first Ace and 3 for the second, giving 12 sequences, while the total sequences of two cards is 52 * 51. The ratio 12 / (52 * 51) reduces to 1 / 221, consistent with the stepwise calculation.

Why Other Options Are Wrong:
51/1221, 42/221 and 52/1245 do not match the exact product of the two conditional probabilities. They arise from incorrect simplification or from counting unordered outcomes instead of ordered draws.

Common Pitfalls:
Learners sometimes forget that the second draw is conditional on the first, and may treat probabilities as independent. Another mistake is to reduce fractions incorrectly or to mix up the number of remaining Aces and remaining cards after the first draw.

Final Answer:
The probability that both cards drawn are Aces is 1/221.