In a binomial distribution, the parameter `n=6`. If `9P(X=4)=P(X=2)`, then p=

Answer

To solve the problem step by step, we start with the given information about the binomial distribution.

### Step 1: Write down the given information
We know that in a binomial distribution, the number of trials \( n = 6 \). We are given that \( 9P(X=4) = P(X=2) \).

### Step 2: Write the expressions for \( P(X=4) \) and \( P(X=2) \)
The probability mass function for a binomial distribution is given by:
\[ P(X=k) = \binom{n}{k} p^k q^{n-k} \]
where \( q = 1 - p \).

Thus, we can write:
\[ P(X=4) = \binom{6}{4} p^4 q^{6-4} = \binom{6}{4} p^4 q^2 \]
\[ P(X=2) = \binom{6}{2} p^2 q^{6-2} = \binom{6}{2} p^2 q^4 \]

### Step 3: Substitute these expressions into the equation
Substituting into the equation \( 9P(X=4) = P(X=2) \):
\[ 9 \cdot \binom{6}{4} p^4 q^2 = \binom{6}{2} p^2 q^4 \]

### Step 4: Simplify the equation
We know that \( \binom{6}{4} = \binom{6}{2} \) (using the property \( \binom{n}{r} = \binom{n}{n-r} \)), so we can cancel \( \binom{6}{2} \) from both sides:
\[ 9 p^4 q^2 = p^2 q^4 \]

### Step 5: Divide both sides by \( p^2 q^2 \) (assuming \( p \neq 0 \) and \( q \neq 0 \))
\[ 9 p^2 = q^2 \]

### Step 6: Substitute \( q = 1 - p \)
Now, substituting \( q = 1 - p \) into the equation:
\[ 9 p^2 = (1 - p)^2 \]

### Step 7: Expand and rearrange the equation
Expanding the right side:
\[ 9 p^2 = 1 - 2p + p^2 \]
Rearranging gives:
\[ 9 p^2 - p^2 + 2p - 1 = 0 \]
\[ 8 p^2 + 2p - 1 = 0 \]

### Step 8: Use the quadratic formula to solve for \( p \)
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 8 \), \( b = 2 \), and \( c = -1 \).
\[ p = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 8 \cdot (-1)}}{2 \cdot 8} \]
\[ p = \frac{-2 \pm \sqrt{4 + 32}}{16} \]
\[ p = \frac{-2 \pm \sqrt{36}}{16} \]
\[ p = \frac{-2 \pm 6}{16} \]

### Step 9: Calculate the two possible values for \( p \)
Calculating the two possible values:
1. \( p = \frac{4}{16} = \frac{1}{4} \)
2. \( p = \frac{-8}{16} = -\frac{1}{2} \)

### Step 10: Determine the valid probability
Since probabilities cannot be negative, we reject \( p = -\frac{1}{2} \). Thus, the valid solution is:
\[ p = \frac{1}{4} \]

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

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