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)`

Answer

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 of getting exactly \( r \) successes in \( n \) trials in a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \( q = 1 - p \).

2. **Set Up the Given Condition**:
According to the problem, we have:
\[ 9P(X=4) = P(X=2) \]

3. **Calculate \( P(X=4) \)**:
Using the binomial probability formula:
\[ P(X=4) = \binom{6}{4} p^4 q^{2} \]
where \( \binom{6}{4} = \frac{6!}{4! \cdot 2!} = 15 \). Thus,
\[ P(X=4) = 15 p^4 (1-p)^2 \]

4. **Calculate \( P(X=2) \)**:
Similarly, for \( P(X=2) \):
\[ P(X=2) = \binom{6}{2} p^2 q^{4} \]
where \( \binom{6}{2} = \frac{6!}{2! \cdot 4!} = 15 \). Thus,
\[ P(X=2) = 15 p^2 (1-p)^4 \]

5. **Substitute into the Condition**:
Now substituting these into the condition:
\[ 9(15 p^4 (1-p)^2) = 15 p^2 (1-p)^4 \]
Simplifying this gives:
\[ 135 p^4 (1-p)^2 = 15 p^2 (1-p)^4 \]

6. **Divide by 15**:
Dividing both sides by 15:
\[ 9 p^4 (1-p)^2 = p^2 (1-p)^4 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ 9 p^4 (1-p)^2 - p^2 (1-p)^4 = 0 \]
Factoring out \( p^2 (1-p)^2 \):
\[ p^2 (1-p)^2 (9p^2 - (1-p)^2) = 0 \]

8. **Solve for \( p \)**:
The equation \( p^2 (1-p)^2 = 0 \) gives \( p = 0 \) or \( p = 1 \), which are not valid since \( p \) must be between 0 and 1. Now solving:
\[ 9p^2 - (1 - 2p + p^2) = 0 \]
Simplifying gives:
\[ 8p^2 + 2p - 1 = 0 \]

9. **Using the Quadratic Formula**:
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 8, b = 2, c = -1 \):
\[ p = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8} \]
\[ p = \frac{-2 \pm \sqrt{4 + 32}}{16} \]
\[ p = \frac{-2 \pm \sqrt{36}}{16} \]
\[ p = \frac{-2 \pm 6}{16} \]
This gives:
\[ p = \frac{4}{16} = \frac{1}{4} \quad \text{or} \quad p = \frac{-8}{16} = -\frac{1}{2} \text{ (not valid)} \]

10. **Final Answer**:
Thus, the value of \( p \) is:
\[ p = \frac{1}{4} \]