One in 9 ships is likely to be wrecked when they are set on sail. When 6 ships are set on sail, the probability for exactly 3 to arrive safely is

Options

A. `(25xx8^(3))/(9^(6))`
B. `1-(1)/(9^(6))`
C. `""^(6)C_(3)((8^(3))/(9^(6)))`
D. `""^(6)C_(3)((8^(6))/(9^(3)))`

Answer

To solve the problem step by step, we will use the concept of binomial probability.

### Step 1: Identify the probabilities
We know that the probability of a ship being wrecked (failure) is given as \( \frac{1}{9} \). Therefore, the probability of a ship arriving safely (success) is:
\[ P = 1 - \frac{1}{9} = \frac{8}{9} \]
Let \( Q \) be the probability of failure:
\[ Q = \frac{1}{9} \]

### Step 2: Define the parameters
We are given that:
- Total number of ships, \( N = 6 \)
- We want to find the probability that exactly 3 ships arrive safely, which means \( R = 3 \).

### Step 3: Use the binomial probability formula
The binomial probability formula is given by:
\[ P(X = R) = \binom{N}{R} P^R Q^{N-R} \]
Substituting the values we have:
\[ P(X = 3) = \binom{6}{3} \left(\frac{8}{9}\right)^3 \left(\frac{1}{9}\right)^{6-3} \]

### Step 4: Calculate \( \binom{6}{3} \)
The binomial coefficient \( \binom{6}{3} \) is calculated as:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

### Step 5: Calculate \( P^3 \) and \( Q^{3} \)
Now we calculate \( P^3 \) and \( Q^{3} \):
\[ P^3 = \left(\frac{8}{9}\right)^3 = \frac{512}{729} \]
\[ Q^{3} = \left(\frac{1}{9}\right)^{3} = \frac{1}{729} \]

### Step 6: Substitute values into the formula
Now substituting these values back into the formula:
\[ P(X = 3) = 20 \cdot \frac{512}{729} \cdot \frac{1}{729} \]

### Step 7: Simplify the expression
Calculating the product:
\[ P(X = 3) = 20 \cdot \frac{512}{729^2} = \frac{10240}{531441} \]

### Final Answer
Thus, the probability that exactly 3 ships arrive safely when 6 ships are set on sail is:
\[ P(X = 3) = \frac{10240}{531441} \]