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:

- Probability of occurrence of $$0$$ at odd places is

$$\frac13$$

Hence,

$$P(1_{\text{odd}})=\frac23$$

- Probability of occurrence of $$0$$ at even places is

$$\frac12$$

Hence,

$$P(1_{\text{even}})=\frac12$$

We need the probability that the sequence

$$10$$

is followed by

$$01$$

That is, the 4-bit pattern

$$1001$$

Now this sequence can occur in two ways.

Case 1: Sequence starts at an odd position

Then positions are:

Odd, Even, Odd, Even

Hence probability is

$$\frac23\times\frac12\times\frac13\times\frac12$$

$$=\frac1{18}$$

Case 2: Sequence starts at an even position

Then positions are:

Even, Odd, Even, Odd

Hence probability is

$$\frac12\times\frac13\times\frac12\times\frac23$$

$$=\frac1{18}$$

Therefore total probability is

$$\frac1{18}+\frac1{18}$$

$$=\frac19$$

Hence, the required probability is

$$\boxed{\frac19}$$