College Algebra, Chapter 10, 10.3, Section 10.3, Problem 66

A slot machine has three wheels. Each wheel has 11 positions: a bar and the digits $0,1,2,...,9$. When the handle is pulled, the three wheels spin independently before coming to rest. Find the probability that the wheels stop on the following positions.

a.) Three bars

The probability that the wheels stop on three bars is $
\displaystyle \frac{1}{11} \times \frac{1}{11} \times \frac{1}{11} = \frac{1}{1331}$

b.) The same number on each wheel

We know that the slot machine consists of $0,1,2,3,...,9$ and a bar. So the probability of getting the same number on each wheel is $\displaystyle \frac{1}{11} \times \frac{1}{11} \times \frac{1}{11} = \frac{1}{1331}$

c.) At least one bar

For this event, we apply the compliment to the probability that all numbers are shown is all slots in the machine. We know that the probability of showing a number in the slot is $\displaystyle \frac{10}{11}$. Thus, the probability of showing at least one bar in all three slots is:


$
\begin{equation}
\begin{aligned}

=& 1 - \left( \frac{10}{11} \times \frac{10}{11} \times \frac{10}{11} \right)
\\
\\
=& \frac{331}{1331}
\\
\\
=& 0.248685

\end{aligned}
\end{equation}
$