Chapter 6, Problem 99CP

a.

To determine

Explain the reason for $P\left(1\right)=\frac{1}{6}$, $P\left(2\right)=\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)$, and $P\left(3\right)={\left(\frac{5}{6}\right)}^2\left(\frac{1}{6}\right)$ .

b.

To determine

Calculate the value of $P\left(4\right)$.

Find the $P\left(x\right)$ for any arbitrary positive integer x.