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 🙂
Best Answer
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
$$ \binom4{1,2,1}\cdot\frac3{12}\cdot\frac4{12}\cdot\frac4{12}\cdot\frac5{12}=\frac{4!}{1!2!1!}\cdot\frac5{3\cdot12\cdot12}=\frac5{36}\approx0.1389\;. $$
This is twice the answer provided.