Question

Let a computer program generate only the digits $0$and $1$ to form a string of binary numbers with a 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 $\text{'}10\text{'}$ is followed by $\text{'}01\text{'}$ 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

$\frac{1}{9}$

Explanation for the correct answer:

Finding the probability that $\text{'}10\text{'}$ is followed by $\text{'}01\text{'}$:

Given,

$$\begin{gathered} P(0\text{at even places})=\frac{1}{2} \\ P(0\mathrm{at}\mathrm{odd}\mathrm{places})=\frac{1}{3} \\ P(1\text{at even places})=1-\frac{1}{2}=\frac{1}{2} \\ P(1\text{at odd places})=1-\frac{1}{3}=\frac{2}{3} \end{gathered}$$

There are only two possibilities to arrange $0110$

Possibility $1$: Odd Even Odd Even

Possibility $2$: Even Odd Even Odd

$P$ ($\text{'}10\text{'}$ is followed by $\text{'}01\text{'}$)

$$\begin{gathered} =\left\{\left(\frac{1}{3}*\frac{1}{2}*\frac{2}{3}*\frac{1}{2}\right)+\left(\frac{1}{2}*\frac{2}{3}*\frac{1}{2}*\frac{1}{3}\right)\right\} \\ =\left\{\frac{1}{18}+\frac{1}{18}\right\} \\ =\frac{1}{9} \end{gathered}$$

Hence, option (C) is the correct answer.