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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 :
$\frac{1}{18}$
$\frac{1}{3}$
$\frac{1}{6}$
$\frac{1}{9}$
Solution
$\underset{\text { odd place }}{1} \underset{\text { even place }}{0} \underset{\text { odd place }}{0} \underset{\text { even place }}{1}$
$\underset{\text { even place }}{1} \underset{\text { odd place }}{0} \underset{\text { even place }}{0} \underset{\text { odd place }}{1}$
$\Rightarrow\left(\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{3}\right)+\left(\frac{2}{2} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2}\right)$
$\Rightarrow \, \frac{1}{9}$
