Question
A card is drawn from a well-shuffled standard deck of 52 cards.
Part A: Two cards are drawn without replacement. Given that the first card is an Ace, find the probability that the second card is also an Ace.
Part B: Two cards are drawn with replacement. Find the probability that both cards are Kings.
Part C: Two cards are drawn without replacement. Find the probability that at least one of them is a Diamond.
Solution — Step by Step
These three parts cover the key ideas: conditional probability, independent events, and the complement method.
Part A: Conditional Probability — Without Replacement
Step 1: Understand "without replacement."
After the first card is drawn and NOT put back, the deck has 51 cards left. The first draw changes the composition of the remaining deck.
Step 2: Apply conditional probability.
We want P(second card is Ace | first card was Ace).
After drawing one Ace, there are 3 Aces left in the remaining 51 cards.
P(second Ace | first Ace) = 3/51 = 1/17
Conditional Probability Formula
P(A|B) = P(A ∩ B) / P(B)
Alternatively: P(both Aces) = 4/52 × 3/51 = 12/2652 = 1/221. P(first is Ace) = 4/52 = 1/13. P(second Ace | first Ace) = (1/221) / (1/13) = 13/221 = 1/17 ✓
Part B: Independent Events — With Replacement
Step 1: Understand "with replacement."
The first card is drawn, noted, then put back. The second draw is from the full 52-card deck again. The two draws are independent.
Step 2: Multiply probabilities directly.
P(first card is King) = 4/52 = 1/13 P(second card is King) = 4/52 = 1/13 (same deck, replacement)
P(both Kings) = 1/13 × 1/13 = 1/169
Part C: At Least One Diamond — Complement Method
Step 1: Use the complement.
P(at least one Diamond) = 1 − P(no Diamonds in two draws)
Step 2: Find P(no Diamonds).
Non-diamond cards = 52 − 13 = 39.
P(first is not Diamond) = 39/52 P(second is not Diamond | first is not Diamond) = 38/51
P(no Diamonds) = 39/52 × 38/51 = 1482/2652 = 247/442
Step 3: Apply complement.
P(at least one Diamond) = 1 − 247/442 = 195/442
We can simplify: GCD(195, 442) = 1, so this is already in lowest terms.
Answers
Part A: P(second Ace | first Ace) = 1/17
Part B: P(both Kings, with replacement) = 1/169
Part C: P(at least one Diamond, without replacement) = 195/442
Why This Works
The distinction between with and without replacement is fundamental:
- With replacement: Each draw is independent. Multiply P(A) × P(B) directly.
- Without replacement: Each draw changes the remaining deck. Use conditional probability P(B|A) for the second draw.
The complement method in Part C is almost always faster for "at least one" problems — calculating P(none) is simpler than adding P(exactly one) + P(exactly two).
💡 Expert Tip
For card problems, remember the structure of a 52-card deck: 4 suits (Hearts, Diamonds, Clubs, Spades), 13 cards per suit (Ace through 10, plus Jack, Queen, King). There are 4 Aces, 4 Kings, 4 Queens, 4 Jacks, 13 Hearts, 13 Diamonds — and the King of Hearts is just 1 card. Keeping this structure clear prevents counting errors.
Common Mistake
⚠️ Common Mistake
Mistake: Using 4/52 for the second draw in a without-replacement problem.
If you draw without replacement, the deck has 51 cards after the first draw, not 52. The probability of drawing an Ace second is 3/51 (given the first was an Ace), not 4/52. Using 4/52 again assumes the first card was replaced — it's the with-replacement answer. Read the problem carefully for "with" or "without" replacement.