Calculating Probability of Drawing Two Aces

This problem asks for the probability of drawing two aces in a row from a standard deck of 52 cards, with the important condition that the first card drawn is not replaced.

Drawing cards without replacement means that the total number of cards, and potentially the number of specific cards, changes after each draw. This makes the events dependent.

Understanding Probability Without Replacement

When events are dependent, the outcome of the first event affects the probability of the second event. To find the probability of two dependent events A and B both happening, we use the formula:

\( P(A \text{ and } B) = P(A) \times P(B|A) \)

Where \( P(B|A) \) is the probability of event B happening given that event A has already happened.

Step-by-Step Probability Calculation

Let's apply this to our problem of drawing two aces successively:

1. Event A: Drawing an ace on the first draw.
2. Event B: Drawing an ace on the second draw, given that the first card drawn was an ace and was not replaced.

Probability of Drawing the First Ace

A standard deck has 52 cards. There are 4 aces in the deck.

The probability of drawing an ace on the first draw (Event A) is:

\( P(\text{First card is Ace}) = \frac{\text{Number of Aces}}{\text{Total Number of Cards}} \)

\( P(\text{First card is Ace}) = \frac{4}{52} = \frac{1}{13} \)

Probability of Drawing the Second Ace (Given the First was an Ace)

After drawing one ace and not replacing it, the deck now has:

• Total cards remaining: \( 52 - 1 = 51 \)
• Aces remaining: \( 4 - 1 = 3 \)

The probability of drawing another ace from this modified deck (Event B given A) is:

\( P(\text{Second card is Ace | First card was Ace}) = \frac{\text{Number of remaining Aces}}{\text{Total remaining Cards}} \)

\( P(\text{Second card is Ace | First card was Ace}) = \frac{3}{51} = \frac{1}{17} \)

Combined Probability of Drawing Two Successive Aces

Now, we multiply the probabilities of the two dependent events to find the probability of drawing two aces successively:

\( P(\text{Drawing Two Aces}) = P(\text{First card is Ace}) \times P(\text{Second card is Ace | First card was Ace}) \)

\( P(\text{Drawing Two Aces}) = \frac{1}{13} \times \frac{1}{17} \)

\( P(\text{Drawing Two Aces}) = \frac{1 \times 1}{13 \times 17} \)

\( P(\text{Drawing Two Aces}) = \frac{1}{221} \)

Thus, the probability of drawing two aces successively without replacement from a well-shuffled pack of 52 cards is \( \frac{1}{221} \).

Revision Table: Key Probability Concepts

Additional Information: Deck Composition and Probability

Understanding the composition of a standard 52-card deck is essential for solving card probability problems.

• Total Cards: 52
• Suits: 4 (Hearts, Diamonds, Clubs, Spades)
• Cards per Suit: 13 (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K)
• Total Aces: 4 (one in each suit)
• Total Face Cards (J, Q, K): 12 (3 in each suit)
• Total Number Cards (2-10): 36 (9 in each suit)

When solving probability problems, always identify if the events are independent (with replacement) or dependent (without replacement), as this significantly changes the calculation method for successive events.