5 cards are drawn from a pack of 52 cards. The probability that these 5 will contain just one king is

Options

A. `(.^(48)C_(4))/(.^(52)C_(4))`
B. `(.^(48)C_(4))/(.^(51)C_(4))`
C. `(.^(48)C_(4))/(.^(52)C_(5))xx4`
D. `(.^(48)C_(44))/(.^(51)C_(4))xx4`

Answer

To find the probability that 5 cards drawn from a pack of 52 cards contain exactly one king, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Number of Kings and Non-Kings**:
- In a standard deck of 52 cards, there are 4 kings and 48 non-king cards.

2. **Define the Event**:
- Let \( E \) be the event that exactly one king is drawn when 5 cards are selected.

3. **Choose the King**:
- We need to choose 1 king from the 4 available kings. The number of ways to choose 1 king from 4 is given by the combination:
\[ \binom{4}{1} = 4 \]

4. **Choose the Non-Kings**:
- We also need to choose 4 non-king cards from the remaining 48 non-king cards. The number of ways to choose 4 non-kings from 48 is given by:
\[ \binom{48}{4} \]

5. **Calculate the Total Ways to Choose 5 Cards**:
- The total number of ways to choose any 5 cards from the 52 cards is given by:
\[ \binom{52}{5} \]

6. **Calculate the Probability**:
- The probability \( P(E) \) of drawing exactly one king in 5 cards is given by the ratio of the favorable outcomes to the total outcomes:
\[ P(E) = \frac{\text{Number of ways to choose 1 king and 4 non-kings}}{\text{Total ways to choose 5 cards}} = \frac{\binom{4}{1} \times \binom{48}{4}}{\binom{52}{5}} \]

7. **Substituting the Values**:
- Now we can substitute the values:
\[ P(E) = \frac{4 \times \binom{48}{4}}{\binom{52}{5}} \]

8. **Calculating the Combinations**:
- Calculate \( \binom{48}{4} \) and \( \binom{52}{5} \):
\[ \binom{48}{4} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} = 194580 \]
\[ \binom{52}{5} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2598960 \]

9. **Final Calculation**:
- Substitute back into the probability formula:
\[ P(E) = \frac{4 \times 194580}{2598960} = \frac{778320}{2598960} \]
- Simplifying this fraction gives:
\[ P(E) = \frac{1}{3.34} \approx 0.299 \]

### Final Answer:
The probability that the 5 cards drawn will contain exactly one king is approximately \( 0.299 \) or \( 29.9\% \).