1. Key Concepts and Formulas

• Independent Events: The generation of each digit in the binary string is an independent event. This means that the probability of a specific sequence of digits occurring is the product of the probabilities of each individual digit appearing at its respective position. For a sequence $D_1 D_2 ... D_k$, its probability is $P(D_1) \times P(D_2) \times ... \times P(D_k)$.
• Position-Dependent Probabilities: The core aspect of this problem is that the probability of generating a '0' or a '1' changes based on whether the position in the string is odd or even. Therefore, it's crucial to track the parity (odd or even) of each digit's position within the sequence.
• Averaging over Starting Parity: When the problem does not specify the exact starting position of the target sequence (e.g., "starts at position 1"), but the probabilities are position-dependent, we must consider all possible initial parities for the sequence. Typically, we assume each possible starting parity (odd or even) is equally likely and average the probabilities calculated for each case to find the overall probability.

2. Step-by-Step Solution

Step 1: Define individual probabilities for '0' and '1' at odd and even places. First, let's clearly list the given probabilities and derive the complementary ones. Let $P(D_p)$ denote the probability of digit $D$ occurring at a position with parity $p$.

We are given:

• Probability of '0' at an even place: $P(0_{even}) = P(0_e) = \frac{1}{2}$
• Probability of '0' at an odd place: $P(0_{odd}) = P(0_o) = \frac{1}{3}$

Since any given position can only contain either a '0' or a '1', the probability of '1' at a particular position is $1$ minus the probability of '0' at that same position.

• Probability of '1' at an even place: $P(1_{even}) = P(1_e) = 1 - P(0_e) = 1 - \frac{1}{2} = \frac{1}{2}$
• Probability of '1' at an odd place: $P(1_{odd}) = P(1_o) = 1 - P(0_o) = 1 - \frac{1}{3} = \frac{2}{3}$

Step 2: Identify the target sequence. The problem asks for the probability that '10' is followed by '01'. This implies we are looking for the four-digit binary sequence '1001'. Let's represent this sequence as $D_1 D_2 D_3 D_4$, where $D_1=1$, $D_2=0$, $D_3=0$, and $D_4=1$.

Step 3: Calculate the probability of the sequence '1001' starting at an odd position. Since the probabilities of digits depend on their position's parity, the probability of the sequence '1001' will depend on whether its first digit ($D_1$) falls on an odd or an even position. We consider the case where the sequence starts at an odd position. If $D_1$ is at an odd position, the parities for the subsequent digits will alternate:

• $D_1$ (which is '1') is at an odd position.
• $D_2$ (which is '0') is at an even position.
• $D_3$ (which is '0') is at an odd position.
• $D_4$ (which is '1') is at an even position.

The probability of '1001' occurring starting at an odd position is the product of the individual probabilities for each digit at its respective parity: $P(\text{'1001' starting odd}) = P(D_1=1 \text{ at odd}) \times P(D_2=0 \text{ at even}) \times P(D_3=0 \text{ at odd}) \times P(D_4=1 \text{ at even})$ $P(\text{'1001' starting odd}) = P(1_o) \times P(0_e) \times P(0_o) \times P(1_e)$ Substituting the values calculated in Step 1: $P(\text{'1001' starting odd}) = \left(\frac{2}{3}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times \left(\frac{1}{2}\right)$ $P(\text{'1001' starting odd}) = \frac{2 \times 1 \times 1 \times 1}{3 \times 2 \times 3 \times 2} = \frac{2}{36} = \frac{1}{18}$

Step 4: Calculate the probability of the sequence '1001' starting at an even position. Now, we consider the case where the sequence '1001' starts at an even position. If $D_1$ is at an even position, the parities for the subsequent digits will alternate:

• $D_1$ (which is '1') is at an even position.
• $D_2$ (which is '0') is at an odd position.
• $D_3$ (which is '0') is at an even position.
• $D_4$ (which is '1') is at an odd position.

The probability of '1001' occurring starting at an even position is the product of the individual probabilities for each digit at its respective parity: $P(\text{'1001' starting even}) = P(D_1=1 \text{ at even}) \times P(D_2=0 \text{ at odd}) \times P(D_3=0 \text{ at even}) \times P(D_4=1 \text{ at odd})$ $P(\text{'1001' starting even}) = P(1_e) \times P(0_o) \times P(0_e) \times P(1_o)$ Substituting the values calculated in Step 1: $P(\text{'1001' starting even}) = \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{2}{3}\right)$ $P(\text{'1001' starting even}) = \frac{1 \times 1 \times 1 \times 2}{2 \times 3 \times 2 \times 3} = \frac{2}{36} = \frac{1}{18}$

Step 5: Calculate the overall probability. Since the problem does not specify whether the sequence '1001' starts at an odd or an even position, we assume that it is equally likely to start at either. Therefore, the total probability is the average of the probabilities calculated for these two cases: $P(\text{'1001'}) = \frac{P(\text{'1001' starting odd}) + P(\text{'1001' starting even})}{2}$ $P(\text{'1001'}) = \frac{\frac{1}{18} + \frac{1}{18}}{2}$ $P(\text{'1001'}) = \frac{\frac{2}{18}}{2} = \frac{\frac{1}{9}}{2} = \frac{1}{18}$

3. Common Mistakes & Tips

• Failure to Define All Probabilities: Always start by clearly listing all four individual probabilities: $P(0_o)$, $P(1_o)$, $P(0_e)$, and $P(1_e)$. This systematic approach minimizes calculation errors.
• Ignoring Position Parity: The most frequent error is to overlook the fact that probabilities change based on whether a position is odd or even. Carefully assign the correct probability to each digit in the sequence based on its calculated parity.
• Not Averaging Over Starting Parity: If the problem doesn't specify the starting position's parity, you must consider all possibilities (odd start, even start) and average their probabilities. Failing to do so would only provide a conditional probability, not the overall probability.

4. Summary

This problem required us to calculate the probability of the specific four-digit binary sequence '1001' occurring, given that the probabilities of '0' and '1' are dependent on whether the position is odd or even. We began by establishing all four fundamental probabilities for '0' and '1' at odd and even places. Then, we analyzed two distinct scenarios: the sequence starting at an odd position and the sequence starting at an even position. For each scenario, we multiplied the probabilities of the individual digits based on their specific values and parities. Since the problem did not specify the starting parity, we averaged the probabilities obtained from both scenarios. Both scenarios yielded a probability of $1/18$, resulting in an overall average probability of $1/18$.

5. Final Answer

The probability that '10' is followed by '01' is $\frac{1}{18}$.

The final answer is \boxed{1 \over 18} which corresponds to option (A).