Seven chits are numbered 1 to 7. three are drawn one by one with replacements. The probability that the least number on any selected chit is 5, is

Options

A. `1-(2//7)^(4)`
B. `4.(2//7)^(4)`
C. `(3//7)^(3)`
D. none

Answer

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

**Step-by-Step Solution:**

1. **Identify the Total Outcomes:**
When drawing a chit, there are 7 possible outcomes (the chits numbered 1 to 7). Since we are drawing three chits with replacement, the total number of outcomes when drawing three chits is:
\[ \text{Total Outcomes} = 7 \times 7 \times 7 = 7^3 = 343 \]

**Hint:** Remember that when drawing with replacement, the total outcomes multiply for each draw.

2. **Identify the Favorable Outcomes:**
We need to find the probability that the least number on any selected chit is 5. This means that all drawn chits must be either 5, 6, or 7. The possible numbers that can be drawn are:
- 5
- 6
- 7

Therefore, there are 3 favorable outcomes for each draw (5, 6, or 7).

3. **Calculate the Favorable Outcomes:**
Since we are drawing three chits, and each chit can be either 5, 6, or 7, the number of favorable outcomes is:
\[ \text{Favorable Outcomes} = 3 \times 3 \times 3 = 3^3 = 27 \]

**Hint:** Count the number of options available for each draw and multiply them together.

4. **Calculate the Required Probability:**
The probability that the least number on any selected chit is 5 can be calculated using the formula:
\[ P(\text{least number is 5}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Outcomes}} = \frac{27}{343} \]

**Hint:** Use the probability formula \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} \).

5. **Final Answer:**
Thus, the probability that the least number on any selected chit is 5 is:
\[ \frac{27}{343} \]