(a) Find the probability that in two tosses of a coin, one is heads and one tails. That in six tosses of a die, all six of the faces show up. That in$\mathbf{12}$tosses of a$\mathbf{12}$ -sided die, all$\mathbf{12}$ faces show up. That in n tosses of an n-sided die, all n faces show up.

(b) The last problem in part (a) is equivalent to finding the probability that, when n balls are distributed at random into n boxes, each box contains exactly one ball. Show that for large n, this is approximately${\mathit{e}}^{\mathbf{-}\mathbf{n}}\sqrt{\mathbf{2}\mathbf{\pi }\mathbf{n}}$.

The two tosses of a coin.

Probability is a metric for determining the possibility of an event occurring.

The coin is tossed twice so$N\left(S\right)=4$.

The outcome of two of them contain one head and one tail is$N\left(A\right)=2$.

So, the probability of 2 heads in one or 2 tails in one is given below.

$\begin{array}{c}P\left(A\right)=\frac{N\left(A\right)}{N\left(S\right)}\\ =\frac{1}{2}\end{array}$

The total method of outcome is mentioned below.

$\begin{array}{c}N\left(S\right)=6\times 6\times 6\times 6\times 6\times 6\\ =6^6\end{array}$

Let event B is that all six faces show up.

So, the number of outcomes of B is mentioned below.

$\begin{array}{c}N\left(S\right)=6\times 5\times 4\times 3\times 2\times 1\\ =6!\end{array}$

The probability of B is mentioned above.

$\begin{array}{c}P\left(B\right)=\frac{N\left(B\right)}{N\left(S\right)}\\ =\frac{6!}{6^6}\end{array}$

The formula states that $n!\approx \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$.

The probability becomes as follows.

$\begin{array}{c}P\left(E\right)\approx \frac{\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n}{n^n}\\ \approx \frac{\sqrt{2\pi n}}{e^n}\\ =e^{-n\sqrt{2\pi n}}\end{array}$

Hence, the probability for part (a) is given below.

$\begin{array}{c}P\left(A\right)=\frac{1}{2}\\ P\left(B\right)=\frac{6!}{6^6}\\ P\left(C\right)=\frac{12!}{{12}^{12}}\\ P\left(E\right)=\frac{n!}{n^n}\end{array}$

Part (b) has been proven.