Suppose `X ~ B(n,p) and P(X=3) = P(X=5)` . If `P gt (1)/(2)`

Options

A. ` n le 7`
B. `n gt 8`
C. `n ge 9`
D. `n lt 10`

Answer

To solve the problem, we start with the given information that \( X \) follows a binomial distribution \( X \sim B(n, p) \) and that \( P(X = 3) = P(X = 5) \). We also know that \( p > \frac{1}{2} \).

### Step 1: Write the probability expressions
The probability mass function for a binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Using this, we can write:
\[ P(X = 3) = \binom{n}{3} p^3 (1-p)^{n-3} \]
\[ P(X = 5) = \binom{n}{5} p^5 (1-p)^{n-5} \]

### Step 2: Set the probabilities equal
Since \( P(X = 3) = P(X = 5) \), we have:
\[ \binom{n}{3} p^3 (1-p)^{n-3} = \binom{n}{5} p^5 (1-p)^{n-5} \]

### Step 3: Simplify the equation
We can rearrange the equation:
\[ \frac{\binom{n}{3}}{\binom{n}{5}} = \frac{p^5 (1-p)^{n-5}}{p^3 (1-p)^{n-3}} \]
This simplifies to:
\[ \frac{\binom{n}{3}}{\binom{n}{5}} = \frac{p^2}{(1-p)^2} \]

### Step 4: Calculate the binomial coefficients
The binomial coefficients can be expressed as:
\[ \frac{\binom{n}{3}}{\binom{n}{5}} = \frac{n!/(3!(n-3)!)}{n!/(5!(n-5)!)} = \frac{5! (n-5)!}{3! (n-3)!} = \frac{5 \cdot 4}{2 \cdot 1} = 10 \]
Thus, we have:
\[ 10 = \frac{p^2}{(1-p)^2} \]

### Step 5: Rearranging the equation
Cross-multiplying gives:
\[ 10(1-p)^2 = p^2 \]
Expanding this, we get:
\[ 10(1 - 2p + p^2) = p^2 \]
\[ 10 - 20p + 10p^2 = p^2 \]
\[ 9p^2 - 20p + 10 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ p = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 9 \cdot 10}}{2 \cdot 9} \]
\[ p = \frac{20 \pm \sqrt{400 - 360}}{18} \]
\[ p = \frac{20 \pm \sqrt{40}}{18} \]
\[ p = \frac{20 \pm 2\sqrt{10}}{18} \]
\[ p = \frac{10 \pm \sqrt{10}}{9} \]

### Step 7: Determine the range for \( n \)
Since \( p > \frac{1}{2} \), we need to check which value of \( p \) satisfies this condition.

Calculating \( \frac{10 + \sqrt{10}}{9} \) and \( \frac{10 - \sqrt{10}}{9} \):
- \( \sqrt{10} \approx 3.16 \)
- \( \frac{10 + 3.16}{9} \approx 1.18 \) (not valid since \( p \) must be ≤ 1)
- \( \frac{10 - 3.16}{9} \approx 0.76 \) (valid)

### Step 8: Find \( n \)
We know that:
\[ n - 3 \cdot (n - 4) < 20 \]
This gives us:
\[ n^2 - 7n + 12 < 20 \]
\[ n^2 - 7n - 8 < 0 \]
Factoring gives:
\[ (n - 8)(n + 1) < 0 \]
The solution to this inequality is:
\[ -1 < n < 8 \]
Thus, \( n \leq 7 \).

### Conclusion
The answer is \( n \leq 7 \).