The reasoning developed for counting microstates applies to many other situations involving probability. For example, if you flip a coin 5 times, how many different sequences of 3 heads and 2 tails are possible? Answer: 10 different sequences, such as HTHHT or TTHHH. In contrast, how many different sequences of 5 heads and 0 tails are possible? Obviously only one, HHHHH, and our equation gives $\mathbf{5}\mathbf{!}\mathbf{/}\left[5!0!\right]\mathbf{=}\mathbf{1}$, using the standard definition that 0! is defined to equal 1.

If the coin is equally likely on a single throw to come up heads or tails, any specific sequence like HTHHT or HHHHH is equally likely. However, there is only one way to get HHHHH, while there are 10 ways to get 3 heads and 2 tails, so this is 10times more probable than getting all heads. Use the expression$\mathbf{5}\mathbf{!}\mathbf{/}\left[N!\left(5-N\right)!\right]$to calculate the number of ways to get 0 heads, 1 head, 2 heads, 3 heads, 4 heads, or 5 heads in a sequence of 5 coin tosses. Make a graph of the number of ways vs. the number of heads.

Step by step solution

01

Identification of given data

The number of ways to get head is N.

02

Expression is used to calculate the number of ways.

The expression to calculate the number of ways to get a head is,

$\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{ways}\mathbf{=}\mathbf{5}\mathbf{!}\mathbf{/}\left[N!\left(5-N\right)!\right]$

03

Calculating the number of ways to get head

It is given that the expression to calculate the number of ways to get a head is

$\mathrm{Number}\mathrm{of}\mathrm{ways}=5!/\left[\mathrm{N}!\left(5-\mathrm{N}\right)!\right]$

For the number of ways to get zero head,$\mathrm{N}=0$

Therefore,

$$\begin{gathered} {\mathrm{N}}_0=5!/\left[0!\left(5-0\right)!\right] \\ {\mathrm{N}}_0=1 \end{gathered}$$

The number of ways to get 0 head is 1

For the number of ways to get 1 head, $\mathrm{N}=1$

Therefore,

$$\begin{gathered} {\mathrm{N}}_1=5!/\left[1!\left(5-1\right)!\right] \\ {\mathrm{N}}_1=5!/\left[1!4!\right] \\ {\mathrm{N}}_1=5 \end{gathered}$$

The number of ways to get 1 head is 5

For the number of ways to get 2 heads, $\mathrm{N}=2$

Therefore,

$$\begin{gathered} {\mathrm{N}}_2=5!/\left[2!\left(5-2\right)!\right] \\ {\mathrm{N}}_2=5!/\left[2!3!\right] \\ {\mathrm{N}}_2=10 \end{gathered}$$

The number of ways to get 2 heads is 10

For the number of ways to get 3 heads, $\mathrm{N}=3$

Therefore,

$$\begin{gathered} {\mathrm{N}}_3=5!/\left[3!\left(5-3\right)!\right] \\ {\mathrm{N}}_3=5!/\left[3!2!\right] \\ {\mathrm{N}}_3=10 \end{gathered}$$

The number of ways to get 3 heads is 10

For the number of ways to get 4 heads, $\mathrm{N}=4$

Therefore,

$$\begin{gathered} {\mathrm{N}}_4=5!/\left[4!\left(5-4\right)!\right] \\ {\mathrm{N}}_4=5!/\left[4!1!\right] \\ {\mathrm{N}}_4=5 \end{gathered}$$

The number of ways to get 4 heads is 5

For the number of ways to get 5 heads, $\mathrm{N}=5$

Therefore,

$$\begin{gathered} {\mathrm{N}}_5=5!/\left[5!\left(5-5\right)!\right] \\ {\mathrm{N}}_5=5!/\left[5!0!\right] \\ {\mathrm{N}}_5=1 \end{gathered}$$

The number of ways to get 5 heads is 1

The graph between the number of ways vs the number of the head is shown below:

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