Two cards are drawn one after another.... Two cards are drawn one after another (without replacement) from a well shuffled pack of $52$ playing cards. Show the probability of getting or not getting a face catd by drawing a tree diagram.
My Approach
For first drawn
$$n(S)=52$$
$$n(F)=12$$
$$n(N)=40$$
Then,
$$P(F)=\frac {12}{52}$$
$$P(N)=\frac {40}{52}$$
Note, S=sample space
           F=face cards 
           N= non face cards
What should I do next?
Help please.
Thanks in advance.
Than
 A: Find the probability of getting two non-face cards and subtract from 1.
$$P(\text{Both non-face}) = P(\text{non-face on 1st})P(\text{non-face on 2nd | non-face on first}) = (40/52)(39/51)= 0.5882353.$$
Then
$$P(\text{At least one face}) = 1 - 0.5882353 = 0.4117647.$$
Now, can you figure out the probability of getting exactly one face card?
Here is a simulation in R statistical software of the distribution of
$X = $ number of face cards. It is based on a million performance of the
2 card experiment,
each performance starting with a full shuffled deck. (Expect 2 or 3 place
accuracy.)
m = 10^6; deck = 1:52; n = 2; x = numeric(m)
for(i in 1.:m) {
  draw = sample(deck, n)
  x[i] = sum(draw > 40) }  # face cards numbered 41 thru 52
table(x)/m
x
       0        1        2 
0.587444 0.362590 0.049966 

The random variable $X$ has a 'hypergeometric distribution'. You can see if
that is discussed in your text or look at the Wikipedia article. The histogram below shows the simulated
distribution of $X$ and the dots atop bars show exact hypergeometric
probabilities. (The resolution of the graph on the probability scale
is about two decimal places, so the agreement looks perfect.)

R code for the graph:
hist(x, prob=T, br=(0:3)-.5, col="wheat", main="Face Cards in 2 Draws")
i = 0:2;  pdf = dhyper(i, 12, 40, 2)  # 12 face, 40 non-face, 2 drawn
points(i, pdf, pch=19, col="blue")

A: A Tree diagram looks roughly: 
$$\begin{array}{cc}
  &&\mathsf P(F_1,F_2)
\\ & _{\small\mathsf P(F_2\mid F_1)}\nearrow
\\ & \bullet
\\ & _{\small\mathsf P(N_2\mid F_1)}\searrow
\\ ^{\small\mathsf P(F_1)}\nearrow&& \mathsf P(F_1, N_2)
\\ \circ
\\ ^{\small\mathsf P(N_1)}\searrow&& \mathsf P(N_1, F_2)
\\ & _{\small\mathsf P(F_2\mid N_1)}\nearrow
\\ & \bullet
\\ & ^{\small\mathsf P(N_2\mid N_1)}\searrow
\\ &&\mathsf P(N_1,N_2)
\end{array}$$
Supply the numbers.
