Question

Suppose you flip $50$fair coins.

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

(b) How many ways are there of getting exactly$25$heads and$25$tails?

(c) What is the probability of getting exactly $25$heads and $25$tails?

(d) What is the probability of getting exactly $30$heads and $20$tails?

(e) What is the probability of getting exactly $40$heads and 10 tails?

(f) What is the probability of getting $50$heads and no tails?

(g) Plot a graph of the probability of getting n heads, as a function of n.

Short answer

(a)Possible outcomes (microstates) are there $1.2641\times {10}^{14}\text{microstate,}$

(b) The ways are there of getting exactly $25$heads and$25$tails are $1.2641\times {10}^{14}\text{microstate}$

(c)The is the probability of getting exactly $25$heads and $25$tails are $P=0.1123$

(d) The probability of getting exactly $30$heads and $20$tails? $P=0.04186$

(e) The probability of getting exactly $40$heads and $10$tails? $P=9.1236\times {10}^{-6}$

(f)The probability of getting $50$heads and no tails are $P=8.882\times {10}^{-16}$

(e) A graph of the probability of getting n heads, as a function of n.

Step-by-step solution

Step1:Explanation of 50 fair coins (part a b c and d)

(a) we suppose we flip $50$coins, $n=50$. The total number of microstates is:

$2^n=2^{50}=1.1259\times {10}^{15}\text{microstate}$

(b) The number of microstates receiving $25$heads is as follows:

$\left(\begin{array}{l}N\\ n\end{array}\right)$

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

$\left(\begin{array}{l}50\\ 25\end{array}\right)=\frac{50!}{25!(50-25)!}=1.2641\times {10}^{14}\text{microstate}$

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

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

$\to P=\frac{\left(\begin{array}{l}50\\ 25\end{array}\right)}{2^{50}}=0.1123$

(d) The probability of getting the macrostate of$30$heads and $20$tails, regardless of order, is:

$\to P=\frac{\left(\begin{array}{l}50\\ 30\end{array}\right)}{2^{50}}=\frac{\frac{50!}{30!(50-30)!}}{2^{50}}=0.04186$

Step2:Explanation of  50 fair coins(part e f and  g )

(e)The probability of obtaining the macrostate of$40$heads and$10$tails, regardless of order, is as follows:

localid="1650287550141" $\to P=\frac{\left(\begin{array}{c}50\\ 40\end{array}\right)}{2^{50}}=\frac{\frac{50!}{40!(50-40)!}}{2^{50}}=9.1236\times {10}^{-6}$

(f) Regardless of order, the probability of receiving the macrostate of$50$heads and $0$tails is:

localid="1650287553981" $\to P=\frac{\left(\begin{array}{c}50\\ 0\end{array}\right)}{2^{50}}=\frac{\frac{50!}{0!(50-0)!}}{2^{50}}=8.882\times {10}^{-16}$

(g)The function we need to plot is the number of heads divided by the probability, so is:

localid="1650287557923" $P\left(n\right)=\frac{\mathrm{\Omega }(50,n)}{2^{50}}=\frac{\frac{50!}{n!(50-n)!}}{2^{50}}$

The table shows the values that we intend to plot where n varies from$0$to$50$. I drew the graph in MS Excel, and the image below shows how I generated the data.

Step3:Probability of 50 fair coins

Step4:find graph

The probability decreases sharply as we move away from the midpoint, with an equal number of heads and tails. This is nicely illustrated by a plot of the probability P(N):