Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is selected. Find the probability of selecting a six and then selecting a seven. The probability of selecting a six and then selecting a seven is (Round to three decimal places as needed.)

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**Title: Understanding Probability with Playing Cards**

**Problem Statement:**
Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is selected. Find the probability of selecting a six and then selecting a seven.

**Question:**
- What is the probability of selecting a six and then selecting a seven? 
- The probability should be rounded to three decimal places.

**Solution Explanation:**

To determine the probability of two sequential events where one event affects the other (since the first card is not replaced), use the concept of conditional probability.

### Step-by-Step Calculation:

1. **Probability of Selecting a Six (First Draw):**
   - There are 4 sixes in a standard deck of 52 cards.
   - Thus, the probability of drawing a six is: \( P(6) = \frac{4}{52} \).

2. **Probability of Selecting a Seven (Second Draw):**
   - After drawing a six, there are now 51 cards left in the deck.
   - There are still 4 sevens in the deck.
   - Thus, the probability of drawing a seven after drawing a six is: \( P(7 | 6) = \frac{4}{51} \).

3. **Joint Probability:**
   - The probability of both events occurring in sequence (drawing a six first and then drawing a seven) is the product of the individual probabilities.
   - So, the probability is: \( P(6 \text{ then } 7) = \frac{4}{52} \times \frac{4}{51} \).

### Calculation:
\[ P(6) = \frac{4}{52} = \frac{1}{13} \]
\[ P(7 | 6) = \frac{4}{51} \]

\[ P(6 \text{ then } 7) = \frac{1}{13} \times \frac{4}{51} = \frac{4}{663} \approx 0.006 \]

### Conclusion:
The probability of selecting a six and then selecting a seven, without replacement, rounded to three decimal places is:
\[ 0.006 \]

**Note:** Understanding the role of conditional probability and how the first event affects the second is crucial in solving such problems. This example helps illustrate the concept of dependent events in probability theory.
Transcribed Image Text:**Title: Understanding Probability with Playing Cards** **Problem Statement:** Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is selected. Find the probability of selecting a six and then selecting a seven. **Question:** - What is the probability of selecting a six and then selecting a seven? - The probability should be rounded to three decimal places. **Solution Explanation:** To determine the probability of two sequential events where one event affects the other (since the first card is not replaced), use the concept of conditional probability. ### Step-by-Step Calculation: 1. **Probability of Selecting a Six (First Draw):** - There are 4 sixes in a standard deck of 52 cards. - Thus, the probability of drawing a six is: \( P(6) = \frac{4}{52} \). 2. **Probability of Selecting a Seven (Second Draw):** - After drawing a six, there are now 51 cards left in the deck. - There are still 4 sevens in the deck. - Thus, the probability of drawing a seven after drawing a six is: \( P(7 | 6) = \frac{4}{51} \). 3. **Joint Probability:** - The probability of both events occurring in sequence (drawing a six first and then drawing a seven) is the product of the individual probabilities. - So, the probability is: \( P(6 \text{ then } 7) = \frac{4}{52} \times \frac{4}{51} \). ### Calculation: \[ P(6) = \frac{4}{52} = \frac{1}{13} \] \[ P(7 | 6) = \frac{4}{51} \] \[ P(6 \text{ then } 7) = \frac{1}{13} \times \frac{4}{51} = \frac{4}{663} \approx 0.006 \] ### Conclusion: The probability of selecting a six and then selecting a seven, without replacement, rounded to three decimal places is: \[ 0.006 \] **Note:** Understanding the role of conditional probability and how the first event affects the second is crucial in solving such problems. This example helps illustrate the concept of dependent events in probability theory.
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