If X follows the binomial distribution with parameters n=6 and p and 9p(X=4)=P(X=2), then p is

Options

A. `(1)/(4)`
B. `(1)/(3)`
C. `(1)/(2)`
D. `(2)/(3)`

Explanation

To solve the problem, we need to find the value of \( p \) given that \( X \) follows a binomial distribution with parameters \( n = 6 \) and \( p \), and the condition \( 9P(X = 4) = P(X = 2) \).

### Step-by-Step Solution:

1. **Understand the Binomial Probability Formula**:
The probability mass function for a binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.

2. **Calculate \( P(X = 4) \)**:
For \( n = 6 \) and \( k = 4 \):
\[ P(X = 4) = \binom{6}{4} p^4 (1-p)^{2} \]
The binomial coefficient \( \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15 \).
Therefore:
\[ P(X = 4) = 15 p^4 (1-p)^2 \]

3. **Calculate \( P(X = 2) \)**:
For \( n = 6 \) and \( k = 2 \):
\[ P(X = 2) = \binom{6}{2} p^2 (1-p)^{4} \]
The binomial coefficient \( \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \).
Therefore:
\[ P(X = 2) = 15 p^2 (1-p)^4 \]

4. **Set Up the Equation**:
According to the problem, we have:
\[ 9P(X = 4) = P(X = 2) \]
Substituting the expressions we found:
\[ 9 \times (15 p^4 (1-p)^2) = 15 p^2 (1-p)^4 \]

5. **Simplify the Equation**:
Cancel \( 15 \) from both sides:
\[ 9 p^4 (1-p)^2 = p^2 (1-p)^4 \]
Divide both sides by \( p^2 \) (assuming \( p \neq 0 \)):
\[ 9 p^2 (1-p)^2 = (1-p)^4 \]

6. **Rearranging the Equation**:
Divide both sides by \( (1-p)^2 \) (assuming \( 1-p \neq 0 \)):
\[ 9 p^2 = (1-p)^2 \]

7. **Expand and Rearrange**:
Expanding the right side:
\[ 9 p^2 = 1 - 2p + p^2 \]
Rearranging gives:
\[ 9 p^2 - p^2 + 2p - 1 = 0 \implies 8p^2 + 2p - 1 = 0 \]

8. **Using the Quadratic Formula**:
The quadratic formula is given by:
\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 8 \), \( b = 2 \), and \( c = -1 \):
\[ p = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8} \]
\[ = \frac{-2 \pm \sqrt{4 + 32}}{16} = \frac{-2 \pm \sqrt{36}}{16} = \frac{-2 \pm 6}{16} \]

9. **Calculating the Roots**:
This gives us two potential solutions:
\[ p = \frac{4}{16} = \frac{1}{4} \quad \text{and} \quad p = \frac{-8}{16} = -\frac{1}{2} \]
Since \( p \) must be between 0 and 1, we take:
\[ p = \frac{1}{4} \]

### Final Answer:
The value of \( p \) is \( \frac{1}{4} \).