probability that you get one ace and the rest are face cards if you are dealt with 5 cards from a Standard Deck You are dealt 5 cards from a standard deck. You keep careful track
of the order of the cards you are dealt. What is the probability that you get one ace and the rest are face cards?
Am thinking I can solve the problem as P(AFFFF)+P(FAFFF)+P(FFAFF)+P(FFFAF)+P(FFFFA) Without replacement. But am also wondering if this is a case of a Binomial where n=5 x=1 P=4/52 1-P=48/52
 A: This is not related to a Binomial distribution - which is that of the count of successes in a series of iid Bernoulli trials.  
This senario is more similar to a Hypergeometric Distribution - that of the count of successes in a sample drawn without replacement from a population.   In this case we have two disjoint kinds of favored items in the population (deck).
$$\Pr(\textsf{5 Card Hand = 1 Ace & 4 Face}) ~=~\dfrac{\dbinom{4}{1}\dbinom{12}{4}\color{gainsboro}{\dbinom{52-16}{0}}}{\dbinom{52}{5}\qquad\qquad\quad}$$
A: 
Wondering if this is a case of a Binomial where $n=5,x=1,P=4/52,1-P=48/52$.

No it's not, because $A$ and $F$ are not complementary events.
Keep in mind that some of the cards are neither $A$ nor $F$.
In other words: $P(A)=4/52$, but $P(F)\neq48/52$.
A: There are 4 ace cards, you need 1. There are 12 face cards, you need 4. The number of ways to get 5 cards from 52 is $\binom{52}{5}$.
A: The probability of something happening is its $\frac {\text{successful outcomes}}{\text{possible outcomes}}$
So our successful outcomes for this problem would be $4\cdot 12\cdot 11\cdot 10\cdot 9$ because there are $4$ Aces to choose from, and $12$ face cards to choose from. But once you remove one of the face cards, you're left with $11$, then $10$, and so on...
And we have a total of $52\cdot 51\cdot 50\cdot 49\cdot 48$. So we have: $$P(\text{Choosing Ace and Choosing 4 face cards next})=\frac {4\times 12\times 11\times 10\times 9}{52\times 51\times 50\times 49\times 48}=\boxed{\frac {99}{649,740}}$$
