It is known that at noon at a certain place the sun is hidden by clouds on an average two days out of every three. The chance that the sun will be shinning at noon on at least four out of five specified future days is

Options

A. `1/81`
B. `16/243`
C. `11/243`
D. `7/20`

Answer

To solve the problem, we need to find the probability that the sun will be shining at noon on at least four out of five specified future days.

### Step-by-Step Solution:

1. **Identify the probabilities**:
- The probability that the sun is hidden by clouds is \( P(\text{not shining}) = \frac{2}{3} \).
- The probability that the sun is shining is \( P(\text{shining}) = \frac{1}{3} \).

2. **Define the scenario**:
- We are interested in the probability of the sun shining on at least 4 out of 5 days.

3. **Calculate the probabilities for 4 and 5 days**:
- **Case 1**: The sun shines on exactly 4 days and is hidden on 1 day.
- The number of ways to choose 4 days out of 5 is given by \( \binom{5}{4} = 5 \).
- The probability for this case is:
\[ P(\text{4 shining, 1 not shining}) = \binom{5}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^1 = 5 \cdot \left(\frac{1}{3}\right)^4 \cdot \left(\frac{2}{3}\right) \]

- **Case 2**: The sun shines on all 5 days.
- The number of ways to choose 5 days out of 5 is \( \binom{5}{5} = 1 \).
- The probability for this case is:
\[ P(\text{5 shining}) = \binom{5}{5} \left(\frac{1}{3}\right)^5 = 1 \cdot \left(\frac{1}{3}\right)^5 \]

4. **Combine the probabilities**:
- Now, we add the probabilities from both cases:
\[ P(\text{at least 4 shining}) = P(\text{4 shining, 1 not shining}) + P(\text{5 shining}) \]
\[ = 5 \cdot \left(\frac{1}{3}\right)^4 \cdot \left(\frac{2}{3}\right) + 1 \cdot \left(\frac{1}{3}\right)^5 \]

5. **Calculate each term**:
- For Case 1:
\[ 5 \cdot \left(\frac{1}{3}\right)^4 \cdot \left(\frac{2}{3}\right) = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{10}{243} \]
- For Case 2:
\[ 1 \cdot \left(\frac{1}{3}\right)^5 = \frac{1}{243} \]

6. **Final probability**:
- Now, we add both results:
\[ P(\text{at least 4 shining}) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243} \]

### Conclusion:
The probability that the sun will be shining at noon on at least four out of five specified future days is \( \frac{11}{243} \).