Let a computer program generate only the digits $0$ and $1$ to form a string of binary numbers with probability of occurrence of $0$ at even places be $\frac{1}{2}$ and probability of occurrence of $0$ at the odd place be $\frac{1}{3}.$ Then the probability that $10$ is followed by $01$ is equal to :

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

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

For the digits $\text{'}10\text{'}$ and $\text{'}01\text{'}$ we have the following possibilities 

$\begin{array}{cccc}1& 0& 0& 1\\ odd place& even place& odd place& even place\end{array}$

Or

$\begin{array}{cccc}1& 0& 0& 1\\ even place& odd place& even place& odd place\end{array}$

Thus, the required probability is $\left(\frac{1}{2}·\frac{1}{3}·\frac{1}{2}·\frac{2}{3}\right)+\left(\frac{2}{3}·\frac{1}{2}·\frac{1}{3}·\frac{1}{2}\right)$

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