A man takes a step forward with probability 0.4 and back ward with probability 0.6. The probability that at the end of eleven steps he is one step away from the starting point is:

Options

A. a. `""^(11)C_6 (0.1 )^(11)`
B. b. `""^(11)C_6 (0.24 )^5`
C. c. `""^(11)C_6 (0.2 )^(11)`
D. d. none of these

Answer

To solve the problem step by step, we need to calculate the probability that a man is one step away from the starting point after taking 11 steps, given that he moves forward with a probability of 0.4 and backward with a probability of 0.6.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( p = 0.4 \) (probability of stepping forward).
- Let \( q = 0.6 \) (probability of stepping backward).
- Total steps taken = 11.

2. **Determine the Conditions**:
- We need the man to be one step away from the starting point after 11 steps. This means he can either be:
- 1 step forward (successes) and 10 steps backward (failures).
- 10 steps forward and 1 step backward.
- Let \( x \) be the number of forward steps (successes) and \( y \) be the number of backward steps (failures).
- We know \( x + y = 11 \) and we want \( |x - y| = 1 \).

3. **Set Up the Equations**:
- From \( x + y = 11 \) and \( |x - y| = 1 \), we have two cases:
- Case 1: \( x - y = 1 \) → \( x = y + 1 \)
- Case 2: \( y - x = 1 \) → \( y = x + 1 \)

4. **Solve for Each Case**:
- **Case 1**:
- Substitute \( x = y + 1 \) into \( x + y = 11 \):
\[ (y + 1) + y = 11 \implies 2y + 1 = 11 \implies 2y = 10 \implies y = 5 \implies x = 6 \]
- **Case 2**:
- Substitute \( y = x + 1 \) into \( x + y = 11 \):
\[ x + (x + 1) = 11 \implies 2x + 1 = 11 \implies 2x = 10 \implies x = 5 \implies y = 6 \]

5. **Calculate the Probability**:
- For both cases, we need to calculate the probability of getting 6 successes and 5 failures (or vice versa).
- The probability of getting exactly \( k \) successes in \( n \) trials is given by the binomial formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
- For \( x = 6 \) and \( y = 5 \):
\[ P(6 \text{ successes}, 5 \text{ failures}) = \binom{11}{6} (0.4)^6 (0.6)^5 \]
- For \( x = 5 \) and \( y = 6 \):
\[ P(5 \text{ successes}, 6 \text{ failures}) = \binom{11}{5} (0.4)^5 (0.6)^6 \]

6. **Calculate Each Probability**:
- Calculate \( \binom{11}{6} \) and \( \binom{11}{5} \):
\[ \binom{11}{6} = \frac{11!}{6!5!} = 462 \]
\[ \binom{11}{5} = \frac{11!}{5!6!} = 462 \]
- Now calculate:
\[ P(6, 5) = 462 \cdot (0.4)^6 \cdot (0.6)^5 \]
\[ P(5, 6) = 462 \cdot (0.4)^5 \cdot (0.6)^6 \]

7. **Combine the Probabilities**:
- Total probability:
\[ P(\text{one step away}) = P(6, 5) + P(5, 6) \]

8. **Final Calculation**:
- Substitute the values and calculate the final probability.