An amobeba either splits into two or remains the same or eventually dies out immediately after completion of evary second with probabilities, respectively, 1/2, 1/4 and 1/4. Let the initial amoeba be called as mother amoeba and after every second, the amoeba, if it is distinct from the previous one, be called as 2nd, 3rd,...generations.
The probability that amoeba population will be maximum after completion of 3 s is

Options

A. `1//2^(7)`
B. `1//2^(6)`
C. `1//2^(8)`
D. None of these

Explanation

To solve the problem, we need to determine the probability that the amoeba population is at its maximum after 3 seconds. We will analyze the situation step by step.

### Step 1: Understand the probabilities
The probabilities for the amoeba's behavior are:
- Splits into two: \( P(S) = \frac{1}{2} \)
- Remains the same: \( P(R) = \frac{1}{4} \)
- Dies out: \( P(D) = \frac{1}{4} \)

### Step 2: Determine the conditions for maximum population
To achieve maximum population after 3 seconds, the amoeba must split at each second. This means:
- At \( t = 0 \): 1 amoeba (mother)
- At \( t = 1 \): splits into 2 amoebas
- At \( t = 2 \): each of the 2 amoebas splits into 2, resulting in 4 amoebas
- At \( t = 3 \): each of the 4 amoebas splits into 2, resulting in 8 amoebas

### Step 3: Calculate the probability for each second
1. **At \( t = 1 \)**: The mother amoeba must split.
\[ P(S) = \frac{1}{2} \]

2. **At \( t = 2 \)**: Both amoebas from the first second must split.
\[ P(S) \times P(S) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

3. **At \( t = 3 \)**: All 4 amoebas must split.
\[ P(S) \times P(S) \times P(S) \times P(S) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

### Step 4: Combine the probabilities
The total probability that the amoeba population is maximum after 3 seconds is the product of the probabilities at each second:
\[ P(\text{max population after 3s}) = P(S) \times P(S) \times P(S) \times P(S) = \frac{1}{2} \times \frac{1}{4} \times \frac{1}{16} \]

Calculating this:
\[ = \frac{1}{2} \times \frac{1}{4} \times \frac{1}{16} = \frac{1}{128} \]

### Final Answer
The probability that the amoeba population will be maximum after completion of 3 seconds is:
\[ \frac{1}{128} \]