Question

Short answer

The table showing the number of microstates is shown below in answer.

The probability of getting all three red beans is (b)\(\frac{1}{{27}}\), and the probability of getting 2 green and 1 orange is (c) \(\frac{1}{9}\).

Step-by-step solution

Macrostate and Microstate

In this problem, the changes in macrostate can be predicted by averaging the number of changes in microstates of the given system.

Constructing the table showing the number of microstates and macrostates   

The following table shows the microstates and macrostates.

Macrostate	Possible Microstate-Red (R), Orange (O) and Green (G)			Microstates
3 red, 0 orange, 0 green	RRR			1
2 red, 1 orange, 0 green	RRO	ROR	ORR	3
2 red, 0 orange, 1 green	RRG	RGR	GRR	3
1 red, 2 orange, 0 green	ROO	ORO	OOR	3
1 red, 0 orange, 2 green	RGG	GRG	GGR	3
1 red, 1 orange, 1 green	ROG	RGO	ORG	6
	OGR	GRO	GOR
0 red, 3 orange, 0 green	OOO			1
0 red, 2 orange, 1 green	GOO	OGO	OOG	3
0 red, 1 orange, 2 green	OGG	GOG	GGO	3
0 red, 0 orange, 3 green	GGG			1

Evaluation of probability of obtaining three red beans  

The probability of getting three red beans is calculated below:

\(P = \frac{1}{{27}}\)

Thus, \(\frac{1}{{27}}\) is the required probability.

Evaluation of probability of obtaining two green and one orange  

The probability of getting 2 green and 1 orange is calculated below:

\(\begin{array}{l}P = \frac{3}{{27}}\\P = \frac{1}{9}\end{array}\)

Thus, \(\frac{1}{9}\) is the required probability.