A contest consist of predicting the result win, draw or defeat of 7 football matches. A sent his entry predicting at random. The probability that his entry will contain exactly 4 correct predictions is

Options

A. `8/(3^(7))`
B. `16/(3^(7))`
C. `280/(3^(7))`
D. `560/(3^(7))`

Answer

To find the probability that A's entry will contain exactly 4 correct predictions out of 7 football matches, we can use the binomial probability formula, which is applicable in this scenario as it involves a fixed number of independent trials (the matches) with two possible outcomes (correct prediction or incorrect prediction).

### Step-by-Step Solution:

1. **Identify the parameters:**
- Total number of matches (n) = 7
- Number of correct predictions desired (r) = 4
- Probability of a correct prediction (p) = 1/3 (since there are three outcomes: win, draw, or defeat)
- Probability of an incorrect prediction (q) = 1 - p = 2/3

2. **Use the binomial probability formula:**
The formula for the probability of getting exactly r successes in n trials is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \(\binom{n}{r}\) is the binomial coefficient, calculated as:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

3. **Plug in the values:**
- Here, \(n = 7\), \(r = 4\), \(p = \frac{1}{3}\), and \(q = \frac{2}{3}\).
- Thus, we need to calculate:
\[ P(X = 4) = \binom{7}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^{7-4} \]

4. **Calculate the binomial coefficient:**
\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

5. **Calculate the probabilities:**
- Calculate \(\left(\frac{1}{3}\right)^4 = \frac{1}{81}\)
- Calculate \(\left(\frac{2}{3}\right)^3 = \frac{8}{27}\)

6. **Combine the results:**
\[ P(X = 4) = 35 \cdot \frac{1}{81} \cdot \frac{8}{27} \]

7. **Simplify the expression:**
\[ P(X = 4) = 35 \cdot \frac{8}{2187} = \frac{280}{2187} \]

### Final Answer:
The probability that A's entry will contain exactly 4 correct predictions is:
\[ \frac{280}{2187} \]