Question

Suppose you flip $20$ fair coins.

(a) How many possible outcomes (microstates) are there?

(b) What is the probability of getting the sequence HTHHTTTHTHHHTHHHHTHT (in exactly that order)?

(c) What is the probability of getting $12$ heads and $8$ tails (in any order)?

Short answer

(a)The possible outcomes microstate are$1048576\text{microstates}$

(b) The probability of getting the sequence is$P=9.536\times {10}^{-7}$

(c)The probability of getting 12 heads and 8 tails are$P=0.12$

Step-by-step solution

Step1:possible outcome of macrostate(part a)

(a)We suppose we flip 20 coins, $n=20$. The total number of microstates will be:

Calculation method

$2^n=2^{20}$

$=1048576$microstate

So, the answer is $2^{20}$or 1048576 microstates.

Step2:The probability of sequence(part b)

(b)The likelihood of obtaining any specific sequence of heads and tails (for example HTHHTTTHTHHHTHHHHTHT) is

$P=\frac{\text{Favorable microstate}}{\text{Total number of microstates}}$

$\begin{array}{r}=\frac{1}{2^n}\\ =\frac{1}{2^{20}}\end{array}$

$=\frac{1}{1048576}$

$=9.536\times {10}^{-7}$

As a result, the answer is for the sequence of heads and tails. (for example HTHHTTTHTHHHTHHHHTHT) is $\frac{1}{2^{20}}$or $9.536\times {10}^{-7}$.

Step3:Probability of 12 heads and 8 tails(part c)

(c)Regardless of order, the probability of getting the macrostate of 12 heads and 8 tails is given by:

$P=\frac{\left(\begin{array}{l}N\\ n\end{array}\right)}{\text{Total number of microstates}}$

where n denotes the number of heads and N denotes the number of coins, so:

$P=\frac{\left(\begin{array}{l}20\\ 12\end{array}\right)}{1048576}$

After arranging we get,

$\left(\begin{array}{l}20\\ 12\end{array}\right)=\frac{20!}{12!(20-12)!}=125970$

Finally solution is

$P=\frac{125970}{1048576}=0.12$