A shoping mall is running a scheme: Each packet of detergent SURF contains a coupon which bears letter of the word SURF, if a person buys at least four packets of detergent SURF, and produce all the letters of the word SURF, then he gets one free packet of detergent.
If person buys 8 such packets, then the probability that he gets exactly one free packets is

Options

A. `7//33`
B. `102//495`
C. `13//55`
D. `34//165`

Explanation

To solve the problem, we need to find the probability that a person who buys 8 packets of detergent SURF gets exactly one free packet. Each packet contains a coupon with one of the letters from the word "SURF." To get a free packet, the person must collect at least one of each letter: S, U, R, and F.

### Step-by-Step Solution:

1. **Identify the total number of packets and letters:**
- The person buys 8 packets of detergent.
- Each packet contains one of the letters S, U, R, or F.

2. **Define the variables:**
- Let \( x_1 \) be the number of S's, \( x_2 \) be the number of U's, \( x_3 \) be the number of R's, and \( x_4 \) be the number of F's.
- We have the equation:
\[ x_1 + x_2 + x_3 + x_4 = 8 \]
- Since we need at least one of each letter to qualify for a free packet, we have the constraints:
\[ x_1 \geq 1, \quad x_2 \geq 1, \quad x_3 \geq 1, \quad x_4 \geq 1 \]

3. **Transform the variables:**
- To simplify the problem, we can define new variables:
\[ y_1 = x_1 - 1, \quad y_2 = x_2 - 1, \quad y_3 = x_3 - 1, \quad y_4 = x_4 - 1 \]
- This transforms our equation to:
\[ (y_1 + 1) + (y_2 + 1) + (y_3 + 1) + (y_4 + 1) = 8 \]
- Simplifying gives:
\[ y_1 + y_2 + y_3 + y_4 = 4 \]
- Here, \( y_1, y_2, y_3, y_4 \geq 0 \).

4. **Count the non-negative integral solutions:**
- The number of non-negative integral solutions to the equation \( y_1 + y_2 + y_3 + y_4 = 4 \) can be found using the stars and bars combinatorial method:
\[ \text{Number of solutions} = \binom{n + r - 1}{r - 1} = \binom{4 + 4 - 1}{4 - 1} = \binom{7}{3} = 35 \]

5. **Count the cases for exactly one free packet:**
- To get exactly one free packet, we need to ensure that one letter appears at least twice while the others appear exactly once.
- Let’s assume one of the letters appears twice (say S), then we can set:
\[ x_1 = 2, \quad x_2 = 1, \quad x_3 = 1, \quad x_4 = 1 \]
- The equation becomes:
\[ 2 + 1 + 1 + 1 = 5 \]
- The remaining 3 packets can be any combination of letters, but we need to ensure that we still have at least one of each letter.

6. **Calculate the number of ways to choose which letter appears twice:**
- There are 4 choices (S, U, R, F) for which letter can appear twice.
- For each choice, the remaining letters can be distributed in:
\[ y_1 + y_2 + y_3 + y_4 = 3 \]
- The number of non-negative integral solutions for this is:
\[ \binom{3 + 4 - 1}{4 - 1} = \binom{6}{3} = 20 \]
- Therefore, the total number of favorable outcomes for exactly one free packet is:
\[ 4 \times 20 = 80 \]

7. **Calculate the total outcomes:**
- The total number of ways to choose 8 packets from the 4 letters (with repetitions allowed) is:
\[ 4^8 = 65536 \]

8. **Calculate the probability:**
- Finally, the probability of getting exactly one free packet is:
\[ P(\text{exactly one free packet}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{80}{65536} = \frac{5}{4096} \]

### Final Answer:
The probability that the person gets exactly one free packet is \( \frac{5}{4096} \).