Without Replacement: You shuffle the deck thoroughly, take out three cards. For this particular problem, the question is "What is the probability these cards are all Kings."
With Replacement: Shuffle the deck, pick out one card, record what you got. Then put it back in the deck, shuffle, pick out one card, record what you got. Then put it back in the deck, pick out one card, record what you got. One might then ask for the probability that all three recorded cards were Kings. In the with replacement situation, it is possible, for example, to get the $\spadesuit$ King, or the $\diamondsuit$ Jack more than once.
For solving the "without replacement" problem, here are a couple of ways. There are $\binom{52}{3}$ equally likely ways to choose $3$ cards. There are $\binom{4}{3}$ ways to choose $3$ Kings. So our probability is $\binom{4}{3}/\binom{52}{3}$.
Or else imagine taking out the cards one at a time. The probability the first card taken out was a King is $\frac{4}{52}$. Given that the first card taken out was a King, the probability the second one was is $\frac{3}{51}$, since there are $51$ cards left of which $3$ are Kings. So the probability the first two cards were Kings is $\frac{4}{52}\cdot\frac{3}{51}$. **Given that the first two were Kings, the probability the third is is $\frac{2}{50}$. So the desired probability is $\frac{4}{52}\cdot\frac{3}{51}\cdot \frac{2}{50}$.
Remark: We could solve the same three Kings problem under the "with replacement" condition. (You were not asked to do that,) The second approach we took above yields the answer $\left(\frac{4}{52}\right)^3$. Since we are replacing the card each time and shuffling, the probability of what the "next" card is is not changed by the knowledge that the first card was a King.
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: $1 - 0.6588 = 0.34$.
Best Answer
Hint
By symmetry,
P(last 3 cards are odd numbered) = P(first 3 cards are odd numbered)
You should be able to take it from there !
PS:
I thought this much of a hint was enough, but I see a lot of comments, so for completeness I am giving the answer here. There will be $10$ odd and $10$ even numbers from $1$ through $20$, so
P(last 3 cards are odd numbered) = P(first 3 cards are odd numbered) $= \dfrac{10}{20}\cdot\dfrac9{19}\cdot\dfrac8{18}$