Verified Answer :
- The probability of drawing an ace on the first draw is \(\frac{4}{52} = \frac{1}{13}\).
- After drawing one ace, the remaining deck has \(51\) cards with \(3\) aces left, so the probability of the second draw being an ace is \(\frac{3}{51} = \frac{1}{17}\).
- By the multiplication rule for dependent events, the total probability is \(\frac{1}{13} \times \frac{1}{17} = \frac{1}{221}\).
- The final probability is \(\boxed{\frac{1}{221}}\).
Explanation
- Analyze the problem and data
We are dealing with a standard deck of \(52\) playing cards. Our objective is to find the probability of drawing two cards, one after the other, without replacement, such that both cards are aces. Initially, there are \(4\) aces in the deck. Because we are not replacing the first card before the second draw, the total number of cards and the number of available aces will change after the first draw.
- Calculate the probability of the first draw
The first step is to calculate the probability of drawing an ace on the first draw, denoted as \(P(A_1)\). Since there are \(4\) aces in a deck of \(52\) cards, the probability is:
\[P(A_1) = \frac{4}{52} = \frac{1}{13}\]
- Calculate the conditional probability of the second draw
Once the first card is drawn and it is an ace, we are left with \(51\) cards in the deck, and only \(3\) of those cards are aces. Therefore, 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 final probability
To find the probability that both events occur (drawing an ace on the first draw AND an ace on the second draw), we use the multiplication rule for dependent events:
\[P(A_1 \cap A_2) = P(A_1) \times P(A_2 | A_1)\]
\[P(A_1 \cap A_2) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}\]
The total probability of drawing two aces is \(\frac{1}{221}\).
Examples
This type of probability calculation is essential in card games like Poker or Blackjack, where knowing the likelihood of drawing specific combinations helps players make strategic decisions. Understanding dependent events allows you to calculate the odds of consecutive successes in many real-world scenarios, such as selecting items from a limited inventory without replacement. For instance, if you have a bag of \(52\) mixed marbles and need to pick two specific ones, this method provides the exact probability of your success. \(\frac{4}{52} \times \frac{3}{51} = \frac{1}{221} \approx 0.0045\).