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

Options

A. `n le 7`
B. `n gt 8`
C. `n ge 9`
D. none of these

Answer

To solve the problem where \( X \sim B(n, p) \) and \( P(X=3) = P(X=5) \) with \( p > \frac{1}{2} \), we will follow these steps:

### Step 1: Write the Probability Expressions
The probability mass function for a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]
Thus, we can express the probabilities for \( X = 3 \) and \( X = 5 \):
\[ P(X = 3) = \binom{n}{3} p^3 (1-p)^{n-3} \]
\[ P(X = 5) = \binom{n}{5} p^5 (1-p)^{n-5} \]

### Step 2: Set the Probabilities Equal
Given that \( P(X=3) = P(X=5) \), we can set the two expressions equal to each other:
\[ \binom{n}{3} p^3 (1-p)^{n-3} = \binom{n}{5} p^5 (1-p)^{n-5} \]

### Step 3: Simplify the Equation
We can cancel out common terms from both sides:
\[ \frac{\binom{n}{3}}{\binom{n}{5}} = \frac{p^5 (1-p)^{n-5}}{p^3 (1-p)^{n-3}} \]
This simplifies to:
\[ \frac{\binom{n}{3}}{\binom{n}{5}} = \frac{p^2}{(1-p)^2} \]

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

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{p^2}{(1-p)^2} = 10(n-4)(n-3) \]

### Step 6: Analyze the Condition \( p > \frac{1}{2} \)
Given that \( p > \frac{1}{2} \), we know that \( \frac{p^2}{(1-p)^2} \) will be greater than 1. Therefore, we have:
\[ 10(n-4)(n-3) > 1 \]
This implies:
\[ (n-4)(n-3) > \frac{1}{10} \]

### Step 7: Solve the Inequality
To solve \( (n-4)(n-3) > \frac{1}{10} \), we can analyze the quadratic:
\[ n^2 - 7n + 12 > \frac{1}{10} \]
This can be rearranged to:
\[ 10n^2 - 70n + 120 > 1 \]
\[ 10n^2 - 70n + 119 > 0 \]

### Step 8: Finding Roots
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{70 \pm \sqrt{(-70)^2 - 4 \cdot 10 \cdot 119}}{2 \cdot 10} \]
Calculating the discriminant:
\[ 4900 - 4760 = 140 \]
Thus:
\[ n = \frac{70 \pm \sqrt{140}}{20} \]

### Step 9: Determine Valid Integer Values for \( n \)
The roots will give us two critical points. We can analyze the intervals to find valid integer values for \( n \). Since \( n \) must be a positive integer, we check values starting from 5 upwards.

### Conclusion
After analyzing the inequalities and conditions, we find that \( n \) must be in the range \( 5 \leq n \leq 7 \).