91Ó°ÊÓ

First, we have to calculate the total number of 13-card hands we can get from a standard 52-card deck. This is given by the combination formula: \[ \text{total number of hands} = \binom{52}{13} \]

1. Calculate the number of hands that have the Ace and King of no suit: \[ \text{hands without Ace and King} = \binom{48}{13} \] 2. Subtract the number of hands without Ace and King from the total number of hands: \[ \text{favorable hands} = \binom{52}{13} - \binom{48}{13} \] 3. Calculate the probability of getting at least one Ace and King of the same suit by dividing the number of favorable hands by the total number of hands: \[ P(\text{Ace and King of at least one suit}) = \frac{\binom{52}{13} - \binom{48}{13}}{\binom{52}{13}} \]

1. Calculate the number of ways we can choose 1 denomination out of 13 to have all 4 cards: \[ \text{choose 1 denomination} = \binom{13}{1} \] 2. Calculate the number of ways to choose the remaining 9 cards from the remaining 48 (since we excluded 4 cards from the chosen denomination): \[ \text{choose 9 other cards} = \binom{48}{9} \] 3. Multiply the number of ways to choose 1 denomination and the number of ways to choose the remaining 9 cards to find the number of favorable hands: \[ \text{favorable hands} = \binom{13}{1} \cdot \binom{48}{9} \] 4. Calculate the probability of having all 4 cards of at least one denomination by dividing the number of favorable hands by the total number of hands: \[ P(\text{all 4 cards of at least one denomination}) = \frac{\binom{13}{1} \cdot \binom{48}{9}}{\binom{52}{13}} \] Now that we have the probabilities for (a) and (b), we can provide the final answers: (a) The probability of having the Ace and King of at least one suit in a 13-card hand is \[ P(\text{Ace and King of at least one suit}) = \frac{\binom{52}{13} - \binom{48}{13}}{\binom{52}{13}} \] (b) The probability of having all 4 cards of at least one of the 13 denominations in a 13-card hand is \[ P(\text{all 4 cards of at least one denomination}) = \frac{\binom{13}{1} \cdot \binom{48}{9}}{\binom{52}{13}} \]