Two cards are drawn at random from a standard pack of 52 playing cards without replacement. What is the probability that both drawn cards are aces?

Introduction / Context:
Card-draw probabilities often use combinations since order does not matter for the set of drawn cards. There are 4 aces in a 52-card deck; we want both drawn cards to be aces without replacement.

Given Data / Assumptions:

• Deck size = 52, aces = 4.
• Draw 2 cards without replacement.

Concept / Approach:

• Total unordered draws = C(52,2).
• Favorable draws (both aces) = C(4,2).
• Probability = C(4,2)/C(52,2).

Step-by-Step Solution:

Verification / Alternative check:
Sequential method: P(first ace) = 4/52; P(second ace | first ace) = 3/51; product = (4/52)*(3/51) = 12/2652 = 1/221. Same result.

Why Other Options Are Wrong:

• 2/13 and 2/26 ignore shrinking deck or use wrong counting.
• “None of these” is false because 1/221 is exact.

Common Pitfalls:

• Using replacement probabilities or confusing ordered vs unordered counting.

Final Answer:
1/221