Question

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 ${}^{\prime }{10}^{\prime }$ is followed by ${}^{\prime }{01}^{\prime }$ is equal to :

• A. $\frac{1}{6}$
• B. $\frac{1}{18}$
• C. $\frac{1}{9}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{9}$

$P\left(0\text{at even place}\right)=\frac{1}{2},P\left(0\text{at odd place}\right)=\frac{1}{3},$
$P\left(1\text{at even place}\right)=\frac{1}{2},P\left(1\text{at odd place}\right)=\frac{2}{3},$
Required string will be of the form $X....1001$
Here Starting $1$ can be at odd position or even position.
$\therefore P\left(10\text{is followed by}01\right)$
$=(\frac{2}{3}\times \frac{1}{2}\times \frac{1}{3}\times \frac{1}{2})+(\frac{1}{2}\times \frac{1}{3}\times \frac{1}{2}\times \frac{2}{3})$
$=\frac{1}{18}+\frac{1}{18}$
$=\frac{1}{9}$