Let a computer program generate only the digits 0 and 1 to form a...

Given the probability that the digit $0$ appear at the even place is $\dfrac{1}{2},$ hence the probability that the digit $1$ appear at the even place is $1 - \dfrac{1}{2} = \dfrac{1}{2} .$

And the probability that the digit $0$ appear at the odd place is $\dfrac{1}{3},$ hence the probability that the digit $1$ appear at the odd place is $1 - \dfrac{1}{3} = \dfrac{2}{3} .$

For the digits $' 10 ' and$' 01 '$we have the following possibilities </p><p>$\begin{cases}1 & 0 & 0 & 1 \\ oddplace & evenplace & oddplace & evenplace\end{cases}$</p><p>Or</p><p>$\begin{cases}1 & 0 & 0 & 1 \\ evenplace & oddplace & evenplace & oddplace\end{cases}$</p><p>Thus, the required probability is \left(\dfrac{1}{2} \cdot \dfrac{1}{3} \cdot \dfrac{1}{2} \cdot \dfrac{2}{3}\right) + \left(\dfrac{2}{3} \cdot \dfrac{1}{2} \cdot \dfrac{1}{3} \cdot \dfrac{1}{2}\right)$

$= \dfrac{1}{18} + \dfrac{1}{18} = \dfrac{1}{9} .$