Probability of selecting different colored balls from a basket

Question

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.

1. 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 $\dfrac{6!}{2! \times 2!} = 180$.

2. 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.

Answer 1

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.