Chapter 15, Problem 47P

To determine

To construct: A table that represents the number of microstates corresponding to each macrostate.

To find: The probability of obtaining three heads and three tails. Also, probability of six heads.

Answer to Problem 47P

A table that represents the number of microstates corresponding to each macrostate is shown in Table 1. The probability of the three heads and three tails is $\frac{5}{16}$ . The probability of six heads is $\frac{1}{64}$ .

Explanation of Solution

Given:

Six coins are dropped on the floor after shaking.

Formula used:

A macrostate is defined as the overall property of a system, but it does not specify the details of the system like the order in which head or tails occurs. While microstate defines each sequence of the system, means a detailed description of every element of a system.

If n coins are tossed, then the total microstates will be $2^{\mathrm{n}}$ .

Probability for obtaining the given macrostate is,

  ${\mathrm{P}}_{\mathrm{n}:\mathrm{m}}=\frac{{\mathrm{W}}_{\mathrm{n}:\mathrm{m}}}{{\mathrm{W}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}}$

Where,

  ${\mathrm{W}}_{\mathrm{n}:\mathrm{m}}$ is the number of microstate corresponding to the given macrostate.

  ${\mathrm{W}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}$ is the total number of microstates.

Calculation:

Six coins are thrown. So, total number of microstates will be

  ${\mathrm{W}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=2^6=64$

There will be 7 macrostates for the given case.

The required table is shown below.

Table 1

S.NO	MACROSTATE	NUMBER OF MICROSTATES
1	6 HEADS	$\mathrm{C}{}_6=1$	${\mathrm{W}}_{6:6}$
2	5 HEADS, 1 TAIL	$\mathrm{C}{}_5=6$	${\mathrm{W}}_{6:5}$
3	4 HEADS, 2 TAILS	$\mathrm{C}{}_4=15$	${\mathrm{W}}_{6:4}$
4	3 HEADS, 3 TAILS	$\mathrm{C}{}_3=20$	${\mathrm{W}}_{6:3}$
5	2 HEADS, 4 TAILS	$\mathrm{C}{}_2=15$	${\mathrm{W}}_{6:2}$
6	1 HEAD, 5 TAILS	$\mathrm{C}{}_1=6$	${\mathrm{W}}_{6:1}$
7	6 TAILS	$\mathrm{C}{}_0=1$	${\mathrm{W}}_{6:0}$

The probability of obtaining 3 heads and 3 tails is,

  $\begin{array}{c}{\mathrm{P}}_{6:3}=\frac{{\mathrm{W}}_{6:3}}{{\mathrm{W}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}}\\ =\frac{20}{64}\\ =\frac{5}{16}\end{array}$

The probability of obtaining 6 heads is,

  $\begin{array}{c}{\mathrm{P}}_{6:6}=\frac{{\mathrm{W}}_{6:6}}{{\mathrm{W}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}}\\ =\frac{1}{64}\end{array}$

Conclusion:

Therefore, the required table is shown in Table 1. Also, the probability of three heads and three tails is $\frac{5}{16}$ while the probability of six heads is $\frac{1}{64}$ .