For a binomial variate X if `n=5,` and P(X=1)=8P(X=3), thenp=

Options

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

Answer

To solve the problem, we need to find the value of \( p \) for a binomial random variable \( X \) with \( n = 5 \) given that \( P(X = 1) = 8P(X = 3) \).

### 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:
\[ P(X = r) = \binom{n}{r} p^r (1 - p)^{n - r} \]
where \( \binom{n}{r} \) is the binomial coefficient.

2. **Set Up the Given Equation**:
We know that:
\[ P(X = 1) = 8P(X = 3) \]
Using the binomial probability formula, we can write:
\[ P(X = 1) = \binom{5}{1} p^1 (1 - p)^{5 - 1} \]
and
\[ P(X = 3) = \binom{5}{3} p^3 (1 - p)^{5 - 3} \]

3. **Substitute the Values**:
Calculate the binomial coefficients:
\[ \binom{5}{1} = 5 \quad \text{and} \quad \binom{5}{3} = 10 \]
Therefore, we can rewrite the probabilities:
\[ P(X = 1) = 5p(1 - p)^4 \]
\[ P(X = 3) = 10p^3(1 - p)^2 \]

4. **Set Up the Equation**:
Substitute these into the equation:
\[ 5p(1 - p)^4 = 8 \times 10p^3(1 - p)^2 \]
Simplifying gives:
\[ 5p(1 - p)^4 = 80p^3(1 - p)^2 \]

5. **Cancel Common Terms**:
Assuming \( p \neq 0 \) and \( 1 - p \neq 0 \), we can divide both sides by \( p(1 - p)^2 \):
\[ 5(1 - p)^2 = 80p^2 \]

6. **Expand and Rearrange**:
Expanding the left side:
\[ 5(1 - 2p + p^2) = 80p^2 \]
This simplifies to:
\[ 5 - 10p + 5p^2 = 80p^2 \]
Rearranging gives:
\[ 5 - 10p - 75p^2 = 0 \]
or:
\[ 75p^2 + 10p - 5 = 0 \]

7. **Solve the Quadratic Equation**:
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 75, b = 10, c = -5 \):
\[ p = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 75 \cdot (-5)}}{2 \cdot 75} \]
\[ = \frac{-10 \pm \sqrt{100 + 1500}}{150} \]
\[ = \frac{-10 \pm \sqrt{1600}}{150} \]
\[ = \frac{-10 \pm 40}{150} \]

8. **Calculate the Possible Values of \( p \)**:
This gives two potential solutions:
\[ p = \frac{30}{150} = \frac{1}{5} \quad \text{and} \quad p = \frac{-50}{150} \text{ (not valid since } p \text{ must be between 0 and 1)} \]

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