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Let's break down the given problem from the image, which is a probability question about drawing cards from a standard deck.

### Problem Details:
You are asked to compute probabilities for two scenarios when drawing 13 cards from a standard 52-card deck:

#### Part (a):
Compute the probability that the hand contains both the **Ace and King of at least one suit**.

#### Part (b):
Compute the probability that the hand contains **all 4 of at least one of the 13 denominations** (e.g., all four aces, or all four kings, etc.).

---

### Part (a) Solution:
We need to calculate the probability that the 13-card hand contains both the **Ace** and **King** of at least one suit.

#### Approach:
1. **Total possible hands:** 
   The number of ways to choose 13 cards from a deck of 52 is:
   \[
   \binom{52}{13}
   \]
   
2. **Favorable hands:**
   Now, we compute how many hands contain both an Ace and a King of at least one suit.

   - For each suit, the probability that both the Ace and King are present in the hand can be computed by first fixing the Ace and King from that suit (we've chosen 2 specific cards), then choosing 11 more cards from the remaining 50 cards in the deck.
   \[
   \binom{50}{11}
   \]
   
   Since there are 4 suits (hearts, diamonds, clubs, spades), this needs to account for combinations where the Ace and King come from any of these suits, but without counting them twice. Using inclusion-exclusion or approximation techniques can give us the final result.

#### Probability formula (simplified):
The probability \( P(A \cap K) \) for one suit can be approximated by:
\[
P(A \cap K) = \frac{\text{Favorable hands}}{\text{Total hands}} = \frac{4 \times \binom{50}{11}}{\binom{52}{13}}
\]

### Part (b) Solution:
We need to compute the probability that the hand contains **all four cards** of at least one denomination (Ace, King, Queen, etc.).

#### Approach:
1. **Total possible hands:**
   Again, the total number of ways to select 13 cards from 52 is \( \binom{52}{13} \).

2. **Favorable hands:**
   For any given rank (Ace, King, etc.), there are exactly 4 cards. To have all 4 cards of a single denomination in the hand, we fix those 4 cards, and then choose the remaining 9 cards from the remaining 48 cards in the deck.
   \[
   \binom{48}{9}
   \]
   
   Since there are 13 different denominations (Ace through King), we multiply by 13. Thus, the number of favorable hands is:
   \[
   13 \times \binom{48}{9}
   \]

#### Probability formula:
The probability is then:
\[
P(\text{all 4 of a denomination}) = \frac{13 \times \binom{48}{9}}{\binom{52}{13}}
\]

---

Would you like me to compute the exact probabilities or discuss the problem further?

### 5 Related Questions:
1. What is the probability of drawing exactly 2 aces in a hand of 13 cards?
2. How does the probability change if we increase the number of cards drawn from 13 to 15?
3. What is the expected number of kings in a 13-card hand?
4. How can we apply inclusion-exclusion to part (a) to refine the estimate?
5. What is the probability of drawing at least one card of each suit in a hand of 13 cards?

### Tip:
Always use combinations when calculating probabilities involving selection without replacement, as it accounts for the reduction in available items.

A hand of 13 cards is taken from a standard 52-card deck. (a) Compute the probability the hand contains the ace and king of at least one suit. (b) Compute the probability the hand contains all 4 of at least 1 of the 13 denominations.

This solution explores how to calculate the probability of a 13-card hand containing both the ace and king of at least one suit, as well as the probability of having all four cards of a denomination. Using combinatorial methods and probability theory, this problem is suitable for college-level students studying probability.

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