Four letters mailed today each have a `(1)/(3)` probability of arriving in two days sooner. What is the probability that exactly two of the the four letters will arrive in two days or sooner ?

Options

A. `(4)/(81)`
B. `(16)/(81)`
C. `(6)/(27)`
D. `(8)/(27)`

Explanation

To solve the problem, we need to find the probability that exactly two out of four letters arrive in two days or sooner, given that each letter has a probability of \( \frac{1}{3} \) of arriving sooner.

### Step-by-Step Solution:

1. **Identify the probabilities**:
- Probability of a letter arriving in two days or sooner: \( p = \frac{1}{3} \)
- Probability of a letter not arriving in two days or sooner: \( q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \)

2. **Determine the number of letters**:
- Total letters mailed: \( n = 4 \)
- Letters that we want to arrive sooner: \( r = 2 \)

3. **Use the binomial probability formula**:
The probability of exactly \( r \) successes (letters arriving sooner) in \( n \) trials (total letters) is given by the formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \( \binom{n}{r} \) is the number of combinations of \( n \) items taken \( r \) at a time.

4. **Calculate the combinations**:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

5. **Calculate \( p^r \) and \( q^{n-r} \)**:
- \( p^r = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \)
- \( q^{n-r} = \left(\frac{2}{3}\right)^{4-2} = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \)

6. **Combine everything into the formula**:
\[ P(X = 2) = \binom{4}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^{2} \]
\[ P(X = 2) = 6 \times \frac{1}{9} \times \frac{4}{9} \]
\[ P(X = 2) = 6 \times \frac{4}{81} = \frac{24}{81} \]

7. **Simplify the fraction**:
\[ \frac{24}{81} = \frac{8}{27} \]

Thus, the probability that exactly two of the four letters will arrive in two days or sooner is \( \frac{8}{27} \).