Verified Answer :
- The probability of drawing an ace on the first pick is \(\frac{4}{52} = \frac{1}{13}\).
- After one ace is removed, the probability of picking another ace from the remaining \(51\) cards is \(\frac{3}{51} = \frac{1}{17}\).
- The total probability of both events occurring is the product of these individual probabilities: \(\frac{1}{13} \times \frac{1}{17}\).
- The final probability is \(\boxed{\frac{1}{221}}\).
Explanation
- Understand the problem and data analysis
To solve this problem, we start by analyzing the composition of a standard deck of playing cards. A standard deck contains \(52\) cards in total. Among these, there are exactly \(4\) aces. We are drawing two cards sequentially, without replacement, which means the total number of cards in the deck and the composition of the deck change after the first draw. Our goal is to find the probability that both cards drawn are aces.
- Calculate the probability of the first draw being an ace
First, consider the probability of drawing an ace on the first draw. Since there are \(4\) aces out of \(52\) total cards, the probability of the first card being an ace, denoted as \(P(A_1)\), is:
\[P(A_1) = \frac{4}{52} = \frac{1}{13}\]
- Calculate the probability of the second draw being an ace given the first was an ace
Next, we consider the scenario where the first card drawn was an ace. Since we do not replace the first card, the deck now contains \(51\) cards total, and only \(3\) of them are aces. Thus, the conditional probability of drawing a second ace given that the first was an ace, denoted as \(P(A_2 | A_1)\), is:
\[P(A_2 | A_1) = \frac{3}{51} = \frac{1}{17}\]
- Calculate the joint probability and conclude
To find the probability that both events occur (drawing an ace first AND drawing an ace second), we multiply the probability of the first event by the conditional probability of the second event:
\[P(A_1 \cap A_2) = P(A_1) \times P(A_2 | A_1)\]
\[P(A_1 \cap A_2) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}\]
Thus, the probability of drawing two aces is \(\frac{1}{221}\).
Examples
This type of probability calculation is foundational for understanding games of chance, such as poker or bridge, where the outcome depends on the changing composition of the deck. For example, knowing the likelihood of drawing specific combinations can help a player estimate their odds of completing a "strong" hand. In general, it teaches how to account for dependent events where the state of the system is altered by each step of the process.