For a binomial variable X if n=5 and P(X=1)=8P(X=3), then p=

Options

A. `(4)/(5)`
B. `(1)/(5)`
C. `(1)/(3)`
D. `(2)/(3)`

Answer

To solve the problem, we need to find the probability \( p \) given that \( n = 5 \) and \( P(X=1) = 8P(X=3) \).

### Step 1: Write the Binomial Probability Formula
The probability of getting exactly \( k \) successes in \( n \) trials in a binomial distribution is given by the formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.

### Step 2: Calculate \( P(X=1) \) and \( P(X=3) \)
For \( n = 5 \):
- For \( k = 1 \):
\[ P(X=1) = \binom{5}{1} p^1 (1-p)^{5-1} = 5p(1-p)^4 \]

- For \( k = 3 \):
\[ P(X=3) = \binom{5}{3} p^3 (1-p)^{5-3} = 10p^3(1-p)^2 \]

### Step 3: Set Up the Equation
According to the problem, we have:
\[ P(X=1) = 8P(X=3) \]
Substituting the expressions we found:
\[ 5p(1-p)^4 = 8 \cdot 10p^3(1-p)^2 \]
This simplifies to:
\[ 5p(1-p)^4 = 80p^3(1-p)^2 \]

### Step 4: Simplify the Equation
We can divide both sides by \( p \) (assuming \( p \neq 0 \)):
\[ 5(1-p)^4 = 80p^2(1-p)^2 \]
Now, divide both sides by \( (1-p)^2 \) (assuming \( 1-p \neq 0 \)):
\[ 5(1-p)^2 = 80p^2 \]

### Step 5: Expand and Rearrange
Expanding the left side:
\[ 5(1 - 2p + p^2) = 80p^2 \]
This leads to:
\[ 5 - 10p + 5p^2 = 80p^2 \]
Rearranging gives:
\[ 5 - 10p + 5p^2 - 80p^2 = 0 \]
\[ 5 - 10p - 75p^2 = 0 \]
Dividing the entire equation by 5:
\[ 1 - 2p - 15p^2 = 0 \]

### Step 6: Solve the Quadratic Equation
Now we can use the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -15, b = -2, c = 1 \):
\[ p = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-15)(1)}}{2(-15)} \]
\[ p = \frac{2 \pm \sqrt{4 + 60}}{-30} \]
\[ p = \frac{2 \pm \sqrt{64}}{-30} \]
\[ p = \frac{2 \pm 8}{-30} \]

Calculating the two possible values:
1. \( p = \frac{10}{-30} = -\frac{1}{3} \) (not valid since probability cannot be negative)
2. \( p = \frac{-6}{-30} = \frac{1}{5} \)

### Final Answer
Thus, the value of \( p \) is:
\[ p = \frac{1}{5} \]

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