Question on Probability & Distribution Good evening everyone,
I am currently working on a self-study question:

There is 15 marbles in a jar. Six are blue, while the rest are green. Five random marbles are selected. What is the probability that the number of blue marbles selected is more than the number of green marbles selected.

This question had really quizzed me. The following is my best attempt, however still falling short of going anywhere near the right answer.

P(Blue) = 0.4, P(Green) = 0.6. Find P(Blue > Green).

Appreciate any pointer and guidiance on how I can proceed please.
 A: 
There is 15 marbles in a jar. Six are blue, while the rest are green. Five random marbles are selected. What is the probability that the number of blue marbles selected is more than the number of green marbles selected.

As I suggested in comments the key is to set up (define) all the events and variables carefully. Once your notation is properly in place, it's simple.
Start with this:
Let $B$ be the event "a blue marble is drawn" and let $G$ be the event "a green marble is drawn".
A very basic approach:
Let a sequence of $B$'s and $G$'s represent the drawing of the corresponding colored marbles in order. Then list the events, and assign them probabilities (it's not at all onerous, I did the whole thing on paper in about 2 minutes):
\begin{eqnarray}
\text{Event}&\quad &\text{Probability}\\
{BBBBB}& &  \frac{6}{15}\frac{5}{14}\frac{4}{13}\frac{3}{12}\frac{2}{11}\\
{BBBBG}& &  \frac{6}{15}\frac{5}{14}\frac{4}{13}\frac{3}{12}\frac{9}{11}\\
{BBBGB}& &  \frac{6}{15}\frac{5}{14}\frac{4}{13}\frac{9}{12}\frac{3}{11}\\
\vdots & &  \vdots\\
{GBGBB}& &  \frac{9}{15}\frac{6}{14}\frac{8}{13}\frac{5}{12}\frac{4}{11}\\
{GGBBB}& &  \frac{9}{15}\frac{8}{14}\frac{6}{13}\frac{5}{12}\frac{4}{11}
\end{eqnarray}
Less basic approach (less trivial but faster):
Consider the events "5 B's are drawn", "4 B's and a G are drawn", "3 B's and 2 G's are drawn", taking care to count the ways each can happen
Even less basic approach (even less trivial but faster still):
Let $X$ be the number of $B$'s in the five marbles drawn. Use the appropriate distribution for $X$.
You should get used to doing it the most basic way first, then build up. I suggest you do this question all three ways and make sure you get the same answer every time.
A: Let $b = \mbox{no of blue marbles}$. Let $A = 6$, $N = 15$, $n = 5$.
Find $P(b=3) + P(b=4) + P(b=5)$.
Using the formulae: $P(X=k) = C_A^k C_{N-A}^{n-k} / C_N^n$
\begin{align*}
P(X=3) &= C_6^3C_{15-6}^{5-3} / C_{15}^5 = 0.23976 \\
P(X=4) &= C_6^4C_{15-6}^{5-4} / C_{15}^5 = 0.044955 \\
P(X=5) &= C_6^5C_{15-6}^{5-5} / C_{15}^5 = 0.001998
\end{align*}
Therefore, $P(b=3) + P(b=4) + P(b=5) = 0.23976 + 0.044955 + 0.001998 = 0.286713$.
