# 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.

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 $$^2C1 \times ^3C2 \times ^4C2 \times ^5C1 = 180$$.

$$\text{Probability(1 red, 2 blue, 2 purple and 1 green)} = \dfrac{180 \times 180}{14^6}$$

• Are you sure the question says with replacement ? Feb 4 at 12:31
• @trueblueanil Yes Sir. Feb 4 at 12:34
• But then the number of balls given have no relevance, and the probability for each color would just be $1/4$ ! Feb 4 at 12:39
• @trueblueanil Sir, can you please share complete solution? Feb 4 at 12:46
• I have given an answer based on what such questions normally are. Can you please reproduce the exact wording of the question ? Feb 4 at 13:33

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.
• Explain, please, why does this term $\binom{6}{1, 2, 2, 1} = \frac{6!}{1!2!2!1!} = \binom{6}{1}\binom{5}{2}\binom{3}{2}\binom{1}{1}$ need? If it is honestly, I don't understand. Why do fractions have degrees? Why don't the denominator of those fractions decrease: $\left(\frac{2}{14}\right)^{1}\left(\frac{3}{13}\right)^{2}\left(\frac{4}{11}\right)^2\left(\frac{5}{9}\right)^1$? Feb 4 at 14:01
• @VadimChernetsov Choose one of the six positions in the sequence of selections for the red ball, two of the five remaining positions for the blue balls, two of the remaining three positions for the purple ball, and fill the remaining position in the sequence with the green ball. The denominators do not decrease since the selections are with replacement, so there are $14$ choices for each of the six trials. Feb 4 at 14:37