# 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.

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Good evening to you – wolfies Apr 27 '14 at 5:39
Start by defining your events and variables. You have used 'Blue' and 'Green' to mean two quite different things in a single line, which is a recipe for not merely confusion, but disaster. You have almost no hope of getting to the right answer if you don't set your notation for the problem up sensibly. – Glen_b Apr 27 '14 at 6:29
I think the most elementary way to approach it is simply to list all mutually exclusive events that satisfy the condition, and then add their probabilities. You need to be able to do it that way before you start trying to apply anything fancier. – Glen_b Apr 27 '14 at 6:47

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.

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.

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Thanks Glen. I managed to solve this question using the Hypergeometric distribution. Thanks for the enlightenment. – user1275515 Apr 27 '14 at 8:43
Do you want to write it up as an answer? – Glen_b Apr 27 '14 at 9:15
yes, definitely. will do so right away – user1275515 Apr 27 '14 at 9:16

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, $P(b=3) + P(b=4) + P(b=5) = 0.23976 + 0.044955 + 0.001998 = 0.286713$.

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Thanks for pointing out Glen. Was careless there looking at another similar question. Double checked, the workings worked out to be right. =) – user1275515 Apr 27 '14 at 9:24