Probability of selecting different colored balls from a basket A basket contains 2 red, 3 blue, 4 purple, and 5 green balls. 6 balls are selected with replacement. What is the probability that 1 red, 2 blue, 2 purple and 1 green ball will be selected?
My Approach
To solve this problem, we first need to determine the total possible number of favorable cases and then probability of each favorable case.

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*Number of favorable cases: What I mean by favorable cases is any combination that satisfies the problem constraints such as R/B/B/P/P/G. Since there are 6 possible options and 2 colors repeat twice, therefore, the total number of favorable cases are $\dfrac{6!}{2! \times 2!} = 180$.


*Probability of any given favorable case: Since 1 red balls needs to be selected out of 2, the total number of combinations is $^2C1$, similarly for blue balls we have $^3C2$, for purple we have $^4C2$, and finally for green balls we have $^5C1$. The total number of such cases will be $^2C1 \times ^3C2 \times ^4C2 \times ^5C1 = 180$.
$$ \text{Probability(1 red, 2 blue, 2 purple and 1 green)} = \dfrac{180 \times 180}{14^6}$$
Is this the right answer? If not, please share the correct answer.
 A: Your denominator is correct since there are $2 + 3 + 4 + 5 = 14$ balls that could be drawn on each of the six trials.
This is a multinomial distribution problem.  The probability of drawing one red ball, two blue balls, two purple balls, and one green ball in six trials is
$$\binom{6}{1, 2, 2, 1}\left(\frac{2}{14}\right)^{1}\left(\frac{3}{14}\right)^{2}\left(\frac{4}{14}\right)^2\left(\frac{5}{14}\right)^1$$
where
$$\binom{6}{1, 2, 2, 1} = \frac{6!}{1!2!2!1!} = \binom{6}{1}\binom{5}{2}\binom{3}{2}\binom{1}{1}$$
is the number of sequences of length six containing one red ball, two blue balls, two purple balls, and one green ball, $(2/14)^1$ is the probability of selecting one red ball, $(3/14)^2$ is the probability of selecting two blue balls, $(4/14)^2$ is the probability of selecting two purple balls, and $(5/14)^1$ is the probability of selecting one green ball.
Your answer is too small since the balls are selected without replacement.  Therefore, there are two possible ways to select a red ball on each trial, three possible ways to select a blue ball on each trial, four possible ways to select a purple ball on each trial, and five possible ways to select a green ball on each trial.  When you use combinations, you are selecting without replacement.
