# Probability that $1$ red ball, $2$ green balls, and $2$ yellow balls will be selected

Question: A bucket contains $4$ red balls, $8$ green balls, and $2$ yellow balls. Five balls are selected randomly with replacement. What is the probability that $1$ red ball, $2$ green balls, and $2$ yellow balls will be selected?

I think the probability of selecting each color does not change since the selection is with replacement. I do not think I can use the binomial distribution for this because it seems off. How can I solve this if the binomial is not working?

Yes, the probabilities are constant between trials and the trials are independent. Notice that this follows the multinomial distribution, with $n = 5$, $x_1 = 1$, $x_2 = 2$, $x_3 = 2$, $p_1 = \frac{4}{14}$, $p_2 = \frac{8}{14}$, $p_3 = \frac{2}{14}$, where the indices $1, 2,$ and $3$ represent red, green and yellow respectively. Then the probability we seek is $$\frac{n!}{x_1!x_2!x_3!}p_1^{x_1}p_2^{x_2}p_3^{x_3} = \frac{5!}{1!\cdot 2!\cdot2!} \left(\frac{4}{14}\right)^1\left(\frac{8}{14}\right)^2\left(\frac{2}{14}\right)^2\approx 0.0571.$$