Key Concepts and Formulas

• Binomial Probability Distribution: This distribution is used for a sequence of independent trials (called Bernoulli trials) where each trial has only two possible outcomes (success or failure), and the probability of success remains constant across trials.

  • In this problem:
    • Trial: Drawing a single card.
    • Success: Drawing an Ace.
    • Failure: Drawing a Non-Ace.
    • Number of trials (n): 2 (since two cards are drawn).
    • Random Variable (X): The number of aces obtained in $n$ trials.
  • The crucial condition for using binomial distribution is that the trials must be independent. This is satisfied because cards are drawn with replacement.

• Binomial Probability Formula: The probability of getting exactly $k$ successes in $n$ independent Bernoulli trials is given by: $P(X=k) = \binom{n}{k} p^k q^{n-k}$ where:

  • $n$ is the total number of trials.
  • $k$ is the number of successes desired.
  • $p$ is the probability of success in a single trial.
  • $q$ is the probability of failure in a single trial, and $q = 1 - p$.
  • $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient, which represents the number of distinct ways to achieve $k$ successes in $n$ trials.

• Basic Probability Formula: The probability of an event is calculated as: $P(\text{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$

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Step-by-Step Solution

Step 1: Determine Basic Probabilities for a Single Draw and Identify Binomial Parameters

First, we need to find the probability of drawing an Ace (success) and a Non-Ace (failure) from a standard 52-card deck for a single draw. This will give us our $p$ and $q$ values.

• Total number of cards in the deck = 52.
• Number of Aces = 4.
• Number of Non-Aces = $52 - 4 = 48$.

Now, we calculate the probabilities for a single draw:

• Probability of drawing an Ace (p): This is our probability of success. $p = P(\text{Ace}) = \frac{\text{Number of Aces}}{\text{Total cards}} = \frac{4}{52} = \frac{1}{13}$
• Probability of drawing a Non-Ace (q): This is our probability of failure. $q = P(\text{Non-Ace}) = \frac{\text{Number of Non-Aces}}{\text{Total cards}} = \frac{48}{52} = \frac{12}{13}$
  • Verification: $p + q = \frac{1}{13} + \frac{12}{13} = 1$, which is correct.

For the binomial distribution, we also identify:

• Number of trials (n): We draw two cards, so $n=2$.
• Random Variable (X): Number of aces obtained. We need to find $P(X=1)$ and $P(X=2)$.

Step 2: Calculate P(X = 1) - Probability of Exactly One Ace

We want to find the probability of getting exactly one Ace in two draws. Here, $n=2$, $k=1$, $p=\frac{1}{13}$, and $q=\frac{12}{13}$. Using the binomial probability formula: $P(X=1) = \binom{2}{1} p^1 q^{2-1} = \binom{2}{1} p^1 q^1$ Let's compute the binomial coefficient: $\binom{2}{1} = \frac{2!}{1!(2-1)!} = \frac{2!}{1!1!} = \frac{2 \times 1}{1 \times 1} = 2$ This coefficient indicates there are 2 ways to get one Ace in two draws (Ace then Non-Ace, or Non-Ace then Ace).

Now, substitute the values into the formula: $P(X=1) = 2 \times \left(\frac{1}{13}\right)^1 \times \left(\frac{12}{13}\right)^1$ $P(X=1) = 2 \times \frac{1}{13} \times \frac{12}{13}$ $P(X=1) = \frac{24}{169}$

• Alternatively (without the binomial formula, for conceptual understanding): To get exactly one Ace, the possible sequences are (Ace on 1st, Non-Ace on 2nd) or (Non-Ace on 1st, Ace on 2nd). Since draws are independent:
  • $P(\text{Ace, Non-Ace}) = P(\text{Ace}) \times P(\text{Non-Ace}) = \frac{1}{13} \times \frac{12}{13} = \frac{12}{169}$
  • $P(\text{Non-Ace, Ace}) = P(\text{Non-Ace}) \times P(\text{Ace}) = \frac{12}{13} \times \frac{1}{13} = \frac{12}{169}$ Since these are mutually exclusive events, we sum their probabilities: $P(X=1) = \frac{12}{169} + \frac{12}{169} = \frac{24}{169}$ Both methods yield the same result.

Step 3: Calculate P(X = 2) - Probability of Exactly Two Aces

We want to find the probability of getting exactly two Aces in two draws. Here, $n=2$, $k=2$, $p=\frac{1}{13}$, and $q=\frac{12}{13}$. Using the binomial probability formula: $P(X=2) = \binom{2}{2} p^2 q^{2-2} = \binom{2}{2} p^2 q^0$ Let's compute the binomial coefficient: $\binom{2}{2} = \frac{2!}{2!(2-2)!} = \frac{2!}{2!0!} = \frac{2}{2 \times 1} = 1$ This coefficient indicates there is only 1 way to get two Aces in two draws (Ace then Ace). Note that $0! = 1$.

Now, substitute the values into the formula: $P(X=2) = 1 \times \left(\frac{1}{13}\right)^2 \times \left(\frac{12}{13}\right)^0$ $P(X=2) = 1 \times \frac{1}{169} \times 1$ $P(X=2) = \frac{1}{169}$

• Alternatively (without the binomial formula, for conceptual understanding): To get exactly two Aces, the only possible sequence is (Ace on 1st, Ace on 2nd). Since draws are independent: $P(\text{Ace, Ace}) = P(\text{Ace}) \times P(\text{Ace}) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$ Both methods yield the same result.

Step 4: Sum the Probabilities P(X = 1) + P(X = 2)

The problem asks for the sum $P(X=1) + P(X=2)$. We add the probabilities calculated in Step 2 and Step 3: $P(X=1) + P(X=2) = \frac{24}{169} + \frac{1}{169}$ $P(X=1) + P(X=2) = \frac{24+1}{169}$ $P(X=1) + P(X=2) = \frac{25}{169}$

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Common Mistakes & Tips

• "With Replacement" vs. "Without Replacement": Always check this crucial detail. "With replacement" implies independence, making binomial distribution applicable. "Without replacement" implies dependence, requiring a different approach (like hypergeometric distribution or conditional probability).
• Understanding $\binom{n}{k}$: Remember that the binomial coefficient accounts for all the different orders in which $k$ successes can occur within $n$ trials. If you're calculating probabilities of specific sequences (like P(A,N)), you need to sum them up. The binomial formula simplifies this by calculating the sum directly.
• Simplifying Fractions Early: Reducing fractions like $\frac{4}{52}$ to $\frac{1}{13}$ at the beginning makes subsequent calculations much simpler and reduces the chance of arithmetic errors.

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Summary

This problem required us to calculate the probability of obtaining at least one Ace when drawing two cards successively with replacement. By identifying the problem as a binomial distribution scenario due to independent trials (with replacement), we first determined the probability of drawing an Ace ($p = 1/13$) and a Non-Ace ($q = 12/13$). We then used the binomial probability formula to calculate $P(X=1)$ (exactly one Ace) and $P(X=2)$ (exactly two Aces) and summed these probabilities to find the required result. The sum $P(X=1) + P(X=2)$ evaluates to $\frac{25}{169}$.

The final answer is \boxed{\text{25 \over 169}}, which corresponds to option (A).