[Math] Bag of 24 distinct objects, four colors, six objects per color, select three (probability).

combinatoricsprobability

Could someone confirm my combinatorics solutions for this question?

Question:

A bag holds 24 different objects, of which 6 are orange, 6 are white, 6 are yellow, and 6 are red. If a juggler selects three objects to juggle, what is the probability that:

a) All three objects are the same color?

b) All three objects are different colors?

Part (a) Solution:

  1. There are a total of 24 distinct objects and three of them are to be selected, so the sample space is $C(24,3)$.
  2. Since all three objects to be picked are to be of the same color, one out of four colors must be selected, $C(4,1)$.
  3. Now three objects must be selected from the six available objects of that color, $C(6,3)$.

Part (a) Answer: $$P = \frac{C(4,1) \cdot C(6,3)}{C(24,3)}$$

Part (b) Solution:

  1. There are a total of 24 distinct objects and three of them are to be selected, so the sample space is $C(24,3)$.
  2. Three colors must be picked and there are four choices for the first color, $C(4,1)$.
  3. There are six balls of that color to choose from and only one must be picked, $C(6,1)$.
  4. There are three choices for the second color, $C(3,1)$, and six balls to choose from, $C(6,1)$.
  5. There are two choices for the third color, $(2,1)$ and six balls to choose from, $C(6,1)$.

Part (b) Answer: $$P = \frac{C(4,1) \cdot C(6,1) + C(3,1) \cdot C(6,1) + C(2,1) \cdot C(6,1)}{C(24,3)}$$

I'm definitely unsure about the (b) solution.

Best Answer

What is the probability that all three objects are the same color.

Your answer is correct.

Alternatively, we choose a ball. The probability that the second ball is the same color as the first is $\frac{5}{23}$. The probability that the third ball is the same color as the first two is $\frac{4}{22}$. Hence, the probability is $$p = 1 \cdot \frac{5}{23} \cdot \frac{4}{22}$$

What is the probability that all three objects are different colors.

Your denominator is correct. We choose three of the four colors, then choose one object from each of the selected colors. Hence, the probability is $$p = \frac{\dbinom{4}{3}\dbinom{6}{1}^3}{\dbinom{24}{3}}$$

Alternatively, we choose a ball. The probability that the second ball we select is of a different color than the first is $\frac{18}{23}$. The probability that the third ball we select is of a different color than the first two is $\frac{12}{22}$. Hence, the probability is $$p = 1 \cdot \frac{18}{23} \cdot \frac{12}{22}$$

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