Seven chits are numbered 1 to 7. Four chits are drawn one by one with replacment. The probability that the least number appearing on any selected chit is 5 is :

Options

A. `((3)/(7))^4`
B. `((6)/(7))^3`
C. `(5xx4xx3)/(7^3)`
D. `((3)/(4))^4`

Answer

To solve the problem, we need to find the probability that the least number appearing on any selected chit is 5 when drawing four chits numbered from 1 to 7 with replacement.

### Step-by-Step Solution:

1. **Understanding the Condition**: The condition states that the least number appearing on any selected chit is 5. This means that all the drawn chits must be either 5, 6, or 7.

2. **Identifying Favorable Outcomes**: Since we can draw chits numbered 5, 6, or 7, we have three options for each draw. Therefore, for four draws, the number of favorable outcomes where all chits are either 5, 6, or 7 is:
\[ \text{Favorable outcomes} = 3^4 \]

3. **Calculating Favorable Outcomes**: We calculate \(3^4\):
\[ 3^4 = 81 \]

4. **Calculating Total Outcomes**: Since there are 7 chits and we are drawing 4 chits with replacement, the total number of outcomes is:
\[ \text{Total outcomes} = 7^4 \]

5. **Calculating Total Outcomes**: We calculate \(7^4\):
\[ 7^4 = 2401 \]

6. **Calculating Probability**: The probability \(P\) that the least number appearing on any selected chit is 5 can be calculated using the formula:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{3^4}{7^4} \]
Substituting the values we calculated:
\[ P = \frac{81}{2401} \]

7. **Final Answer**: Thus, the probability that the least number appearing on any selected chit is 5 is:
\[ P = \frac{81}{2401} \]