P.M.F and C.D.F of a bag of marbles A bag contains $7$ Blue and $3$ red marbles. $Two$ marbles are pulled out of the the bag $without$ replacement. Define $X$ as the number of $Red$ marbles out.
a) Determine the probability mass function $f(x)$:
b)Determine the cumulative distribution function $F(x)$ 
c) Determine the probability of pulling out at most one $Blue$ marble.
d) Compute the mean of $X$
 A: The hypergeometric PMF of the number $X$ of red marbles in two draws without without replacement from a bag of 10 marbles of which 3 are red is as follows:
$$P(X = k) = \frac{{3 \choose k}{7 \choose 2 - k}}{{10 \choose 2}}.$$
for $k = 0, 1 ,2.$
From R statistical software we have the following PMF table and bar chart:
k = 0:2;  pmf = dhyper(k, 3, 7, 2)
cbind(k, pmf)
  k        pmf
[1,] 0 0.46666667
[2,] 1 0.46666667
[3,] 2 0.06666667

These are the same as the values $P(X = 0) = P(X = 1) = 21/45 = 7/15$ and
$P(X = 2) = 3/45 = 1/15$ that you give in your Comment. So you are on the right
track so far.

The CDF of this random variable is defined as $F_X(x) = P(X \le x),$ for all real numbers $x.$
This is a 'stair-step' function with jumps of $7/15$ at $x = 0$ and $x = 1$ and
a jump of $1/15$ at $x = 2.$ Technically (because of the $\le$-sign in the definition), the CDF takes the upper value at each jump (illustrated with heavy black dots).
Although the 'interesting part' of the CDF is in the interval $[0,2],$ notice that
the CDF is defined for all real numbers $x.$ (I have plotted the CDF for values of $x$ in the interval $(-1,3).$

For illustration, let's look at one point on the CDF plot. Take $x = 1.3.$
Then $$F_X(1/3) = P(X \le 1.3) = P(X = 0) + P(X = 1) = 42/45 =14/15.$$ The red dotted
lines in the plot below show the relevant values $x = 1.3$ and $F_X(1.3) = 14/15.$
