a) There are $ \binom{52}{5} = 2,598,960 $ ways of choosing 5 cards. There are $\binom{4}{4} = 1 $ way to select the 4 aces. So there are $\binom{48}{1}= 48 $ ways to select the remaining card. Thus there are a total of 48 ways to select 5 cards such that 4 of them are aces, and the probability is:
$\frac{48}{2,598,960} = \frac{1}{54,145}$.
b) There are $\binom{4}{4} = 1 $ way to choose the 4 aces, and there are $\binom{4}{1} = 4 $ ways to choose a king. So there are $1\times4 = 4 $ ways to choose 5 cards such that 4 are aces and the other is a king card. The probability is: $\frac{4}{2,598,960} = \frac{1}{649,740} $.
c) There are $\binom{4}{3} = 4 $ ways to choose 3 ten cards, and there are $\binom{4}{2} = 6 $ ways of choosing 2 jacks. So there are $ 4\times 6 = 24 $ ways to choose 5 cards such that 3 are ten and 2 are jacks. The probability for this case is: $\frac{24}{2,598,960} = \frac{1}{108,290} $.
d) The probability of 5 non-ace cards is: $\frac{\binom{48}{5}}{\binom{52}{5}} = \frac{1,712,304}{2,598,960} = 0.6588 $,
so the probability of getting 5 card at least one ace is: $1 - 0.6588 = 0.34$.
The probability that you will have at most 3 kings is the probability that you will have less than 4.
$$\mathsf P(K\leq 3) = 1 -\mathsf P(K=4)$$
The probability that you will have exactly all four kings is the count of ways to select 4 kings and 1 other card divided by the count of ways to select any 5 cards.
$$\mathsf P(K=4)~=~\dfrac{{^{4}\mathrm C_{4}}\cdot{^{(52-4)}\mathrm C_{(5-4)}}}{^{52}\mathrm C_5}$$
Put it together.
$$\mathsf P(K\leq 3) ~=~ 1 -\dfrac{{^{4}\mathrm C_{4}}\cdot{^{(52-4)}\mathrm C_{(5-4)}}}{^{52}\mathrm C_5}$$
Best Answer
You either need the suits distributed $4222$ or $3322$. The chance of $4222$ is $$\frac{{4 \choose 1} (\text{suit with four cards}) {13 \choose 4}{13 \choose 2}^3}{52 \choose 10}$$ The chance of $3322$ is $$\frac{{4 \choose 2} (\text{suits with three cards}) {13 \choose 3}^2{13 \choose 2}^2}{52 \choose 10}$$ for a total of $\displaystyle\frac {7592832}{27657385} \approx 0.2745$