# Probability of picking marbles from a bag with only the ratio of marbles given

Here is a question that is puzzling me:

A bag contains a large number of marbles; the numbers of the red, blue and yellow marbles are in the ratio $3:4:5$. Four marbles are randomly drawn without replacement. What is the probability that you will draw $1$ red marble, $2$ blue marbles and $1$ yellow marble?

The answer should be $0.0694$ (could potentially be rounded off a bit).

Okay, here is where I am confused. If they gave us, instead of the ratios that there were $3$ reds, $4$ blues and $5$ yellows and we are drawing without replacement that would be simple, but I am just not sure how to look at this. A fellow classmate said that a clue could be that, because they're in ratio and there's a "large number" of marbles, the probabilities would stay the same even after picking a marble, which I now understand,but still cannot reach the answer using this information.

Thank you in advance for your help :)

• If you had $\{3,4,5\}$ marbles then taking one out changes the probability significantly: The first draw is red with probability $\frac 3{12}$ but given that the second one is again red with probability $\frac 2{11}$. Very different. Now try it with $\{300,400,500\}$ – lulu May 28 '16 at 11:18
• Your fellow classmate is correct. – drhab May 28 '16 at 11:18
• In small sample you are using multihypergeometric to calculate the exact probabilities; in your large sample case, you are using multinomial. – BGM May 28 '16 at 11:49

The answer provided is wrong. In the limit of very many marbles, it doesn't matter whether you draw with or without replacement, since the ratios will stay approximately the same. Thus in this limit the probability of drawing $1$ red, $2$ blue and $1$ yellow marbles is
$$\binom4{1,2,1}\cdot\frac3{12}\cdot\frac4{12}\cdot\frac4{12}\cdot\frac5{12}=\frac{4!}{1!2!1!}\cdot\frac5{3\cdot12\cdot12}=\frac5{36}\approx0.1389\;.$$