From a box containing 20 tickets marked with numbers 1 to 20, four tickets are drawn one by one. After each draw, the ticket is replaced. The probability that the largest value of tickets drawn is 15 is.

Options

A. `((3)/(4))^4`
B. `(27)/(320)`
C. `(27)/(1280)`
D. none of these

Explanation

To solve the problem of finding the probability that the largest value of tickets drawn is 15 when drawing four tickets from a box containing tickets numbered from 1 to 20 (with replacement), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Condition**:
We need to find the probability that the largest ticket drawn is exactly 15. This means that at least one of the tickets must be 15, and all other tickets must be less than or equal to 15.

2. **Identifying Possible Values**:
The tickets can have values from 1 to 20. For the largest ticket to be 15, the possible ticket values can only be from the set {1, 2, 3, ..., 15}.

3. **Calculating Favorable Outcomes**:
- We need to ensure that at least one ticket drawn is 15.
- The other three tickets can be any of the numbers from 1 to 15 (but not greater than 15).
- Therefore, for each of the three tickets that are not 15, we have 15 choices (1 to 15).
- The total number of favorable outcomes when one of the tickets is 15 can occur in four different positions (1st, 2nd, 3rd, or 4th draw).

4. **Calculating Total Outcomes**:
- The total number of outcomes when drawing four tickets from 20 with replacement is \(20^4\).

5. **Calculating the Probability**:
- The number of favorable outcomes where at least one ticket is 15 and the rest are from 1 to 15 is given by:
\[ 4 \times 15^3 \]
- Therefore, the probability \(P\) that the largest ticket drawn is 15 can be calculated as:
\[ P = \frac{4 \times 15^3}{20^4} \]

6. **Simplifying the Probability**:
- We can simplify this expression:
\[ P = \frac{4 \times 15^3}{20^4} = \frac{4}{20^4} \times 15^3 \]
- This can be further simplified:
\[ P = \frac{4}{20^4} \times 15^3 = \frac{4}{(4 \times 5)^4} \times 15^3 = \frac{4}{4^4 \times 5^4} \times 15^3 = \frac{1}{5^4} \times \frac{15^3}{4^3} \]
- Now calculating \(15^3 = 3375\) and \(5^4 = 625\), we find:
\[ P = \frac{3375}{625 \times 64} = \frac{3375}{40000} \]

7. **Final Calculation**:
- After simplifying, we find that:
\[ P = \frac{27}{320} \]

### Conclusion:
Thus, the probability that the largest value of tickets drawn is 15 is \(\frac{27}{320}\).