We are tasked with finding the probability of drawing exactly one King in a hand of seven cards from a deck of 60 cards, which contains 40 Aces and 20 Kings.

Step 1: Total possible hands

The total number of ways to choose 7 cards from a deck of 60 cards can be calculated using combinations:

$\text{Total possible hands} = \binom{60}{7}$

Where $\binom{60}{7}$ represents the number of ways to select 7 cards from 60, calculated as:

$\binom{60}{7} = \frac{60!}{7!(60-7)!}$

Step 2: Number of favorable outcomes

To draw exactly 1 King and 6 Aces, we need to:

1. Choose 1 King from the 20 Kings.
2. Choose 6 Aces from the 40 Aces.

The number of favorable ways to do this can be calculated by multiplying the number of ways to select 1 King and the number of ways to select 6 Aces:

$\text{Favorable outcomes} = \binom{20}{1} \times \binom{40}{6}$

Where:

• $\binom{20}{1} = 20$ (ways to choose 1 King)
• $\binom{40}{6} = \frac{40!}{6!(40-6)!}$ (ways to choose 6 Aces)

Step 3: Probability

The probability is the ratio of favorable outcomes to total possible outcomes:

$P(\text{exactly 1 King}) = \frac{\binom{20}{1} \times \binom{40}{6}}{\binom{60}{7}}$

Step 4: Simplification and Calculation

Let me calculate the values of the combinations to provide the exact probability.

$\binom{20}{1} = 20$

$\binom{40}{6} = \frac{40 \times 39 \times 38 \times 37 \times 36 \times 35}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 3838380$

$\binom{60}{7} = \frac{60 \times 59 \times 58 \times 57 \times 56 \times 55 \times 54}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 386206920$

So, the probability is:

$P(\text{exactly 1 King}) = \frac{20 \times 3838380}{386206920} = \frac{76767600}{386206920} \approx 0.1989$

Thus, the probability of drawing exactly one King in a hand of seven cards is approximately 0.1989, or 19.89%.

Would you like to explore further details or calculations? Here are some related questions you might find interesting:

1. How does the probability change if we draw a different number of cards?
2. How would the probability change if the deck had a different ratio of Kings to Aces?
3. What is the probability of drawing no Kings at all in a hand of 7 cards?
4. How can we calculate the probability of drawing at least one King in a hand of 7 cards?
5. How does the concept of "hypergeometric distribution" apply to this scenario?

Tip: When solving probability problems involving combinations, it's often helpful to recognize when you're dealing with hypergeometric distribution, which is commonly used when sampling without replacement.