Computer is generating binary digits , Probability of '0' at odd position is `1/3` and probability of '0' at even position is `1/2` . Find the probability that 10 is immediately followed by 01

Answer

To solve the problem of finding the probability that the sequence "10" is immediately followed by "01" in a binary digit generation scenario, we can break it down step-by-step.

### Step 1: Understand the Problem
We need to find the probability of the sequence "1001" occurring in the binary digits generated by the computer. The probabilities of generating '0' and '1' at odd and even positions are given as follows:
- Probability of '0' at odd positions = \( \frac{1}{3} \)
- Probability of '1' at odd positions = \( 1 - \frac{1}{3} = \frac{2}{3} \)
- Probability of '0' at even positions = \( \frac{1}{2} \)
- Probability of '1' at even positions = \( 1 - \frac{1}{2} = \frac{1}{2} \)

### Step 2: Identify the Positions
In the sequence "1001":
- The first '1' is at position 1 (odd)
- The first '0' is at position 2 (even)
- The second '0' is at position 3 (odd)
- The second '1' is at position 4 (even)

### Step 3: Calculate the Probability for Each Position
Now we calculate the probability for each digit in the sequence "1001":
1. Probability of '1' at position 1 (odd) = \( \frac{2}{3} \)
2. Probability of '0' at position 2 (even) = \( \frac{1}{2} \)
3. Probability of '0' at position 3 (odd) = \( \frac{1}{3} \)
4. Probability of '1' at position 4 (even) = \( \frac{1}{2} \)

### Step 4: Multiply the Probabilities
Now, we multiply the probabilities of each digit in the sequence "1001":
\[ P(1001) = P(1 \text{ at position 1}) \times P(0 \text{ at position 2}) \times P(0 \text{ at position 3}) \times P(1 \text{ at position 4}) \]
\[ = \left(\frac{2}{3}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times \left(\frac{1}{2}\right) \]

### Step 5: Perform the Multiplication
Calculating the above expression:
\[ = \frac{2}{3} \times \frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} = \frac{2 \times 1 \times 1 \times 1}{3 \times 2 \times 3 \times 2} = \frac{2}{36} = \frac{1}{18} \]

### Step 6: Final Result
Thus, the probability that the sequence "10" is immediately followed by "01" is:
\[ \frac{1}{18} \]