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It seems the specific probability problem you're asking about wasn't included in your message. To help you understand how to solve such problems, let’s walk through a common example that covers key concepts like dependent events and conditional probability:

Example Problem

Question: A standard deck has 52 cards (4 aces). What is the probability of drawing two aces in a row without replacing the first card?

Step-by-Step Analysis & Solution

This problem involves dependent events (the outcome of the first draw affects the second draw). We use the multiplication rule for conditional probability:
[ P(A \cap B) = P(A) \times P(B|A) ]
Where:

• (A): First card is an ace
• (B): Second card is an ace
• (P(B|A)): Probability of (B) given (A) (first card was an ace)

1. Probability of drawing an ace first ((P(A)))

There are 4 aces out of 52 cards:
[ P(A) = \frac{\text{Number of aces}}{\text{Total cards}} = \frac{4}{52} = \frac{1}{13} ]

2. Probability of drawing an ace second ((P(B|A)))

After drawing one ace:

• Remaining aces: (4-1=3)
• Remaining total cards: (52-1=51)

[ P(B|A) = \frac{\text{Remaining aces}}{\text{Remaining cards}} = \frac{3}{51} = \frac{1}{17} ]

3. Joint Probability of Two Aces

Multiply the two probabilities:
[ P(\text{Two aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} \approx 0.45\% ]

Key Takeaways

• For dependent events: Always adjust probabilities based on previous outcomes.
• For independent events (e.g., flipping a coin twice): Multiply the probabilities directly ((P(A) \times P(B))).

If you share the specific problem you’re working on, I can provide a tailored solution!
(\boxed{\frac{1}{221}}) (for the example above)

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