If a binomial distribution have parameters `9, 1//3` then P(X=4)=

Options

A. `(448)/(2187)`
B. `(224)/(1186)`
C. `(112)/(1046)`
D. `(94)/(886)`

Answer

To solve the problem of finding \( P(X = 4) \) for a binomial distribution with parameters \( n = 9 \) and \( p = \frac{1}{3} \), we will use the binomial probability formula:

\[ P(X = r) = \binom{n}{r} p^r q^{n - r} \]

where:
- \( n \) is the number of trials,
- \( r \) is the number of successful trials,
- \( p \) is the probability of success on a single trial,
- \( q \) is the probability of failure on a single trial, which can be calculated as \( q = 1 - p \).

### Step-by-Step Solution:

1. **Identify the parameters**:
- Given \( n = 9 \) and \( p = \frac{1}{3} \).
- Calculate \( q \):
\[ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \]

2. **Set the value of \( r \)**:
- We need to find \( P(X = 4) \), so \( r = 4 \).

3. **Apply the binomial probability formula**:
- Substitute \( n \), \( r \), \( p \), and \( q \) into the formula:
\[ P(X = 4) = \binom{9}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^{9 - 4} \]

4. **Calculate \( \binom{9}{4} \)**:
- The binomial coefficient \( \binom{9}{4} \) can be calculated as:
\[ \binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]

5. **Calculate \( \left(\frac{1}{3}\right)^4 \)**:
- Compute \( \left(\frac{1}{3}\right)^4 \):
\[ \left(\frac{1}{3}\right)^4 = \frac{1}{81} \]

6. **Calculate \( \left(\frac{2}{3}\right)^5 \)**:
- Compute \( \left(\frac{2}{3}\right)^5 \):
\[ \left(\frac{2}{3}\right)^5 = \frac{32}{243} \]

7. **Combine all parts**:
- Now substitute back into the probability formula:
\[ P(X = 4) = 126 \times \frac{1}{81} \times \frac{32}{243} \]

8. **Simplify the expression**:
- First, multiply \( 126 \) and \( 32 \):
\[ 126 \times 32 = 4032 \]
- Now multiply the denominators:
\[ 81 \times 243 = 19683 \]
- Thus, we have:
\[ P(X = 4) = \frac{4032}{19683} \]

### Final Result:
\[ P(X = 4) = \frac{4032}{19683} \]