`X` follows a binomial distribution with parameters `n` and `p` where `0ltPlt1` . If `(P(X=r))/(P(X=n-r))` is independent of `n`and `r` then `p=`

Options

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

Answer

To solve the problem, we need to find the value of \( p \) such that the ratio \( \frac{P(X=r)}{P(X=n-r)} \) is independent of \( n \) and \( r \).

### Step-by-step Solution:

1. **Understand the Binomial Probability Formula**:
The probability mass function for 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, \( p \) is the probability of success, and \( (1-p) \) is the probability of failure.

2. **Write the Probability for \( P(X=n-r) \)**:
Similarly, we can express \( P(X=n-r) \):
\[ P(X = n-r) = \binom{n}{n-r} p^{n-r} (1-p)^{r} \]
Note that \( \binom{n}{n-r} = \binom{n}{r} \).

3. **Set Up the Ratio**:
Now, we can set up the ratio:
\[ \frac{P(X=r)}{P(X=n-r)} = \frac{\binom{n}{r} p^r (1-p)^{n-r}}{\binom{n}{r} p^{n-r} (1-p)^{r}} \]
Since \( \binom{n}{r} \) cancels out, we have:
\[ \frac{P(X=r)}{P(X=n-r)} = \frac{p^r (1-p)^{n-r}}{p^{n-r} (1-p)^{r}} = \frac{p^r}{p^{n-r}} \cdot \frac{(1-p)^{n-r}}{(1-p)^{r}} = \frac{p^r (1-p)^{n-r}}{p^{n-r} (1-p)^{r}} = \left(\frac{p}{1-p}\right)^{r} \cdot (1-p)^{n-2r} \]

4. **Independence Condition**:
For the ratio to be independent of \( n \) and \( r \), the expression \( (1-p)^{n-2r} \) must not depend on \( n \) or \( r \). This can only happen if the exponent \( n-2r = 0 \) or if \( \frac{1-p}{p} = 1 \).

5. **Solve for \( p \)**:
Setting \( \frac{1-p}{p} = 1 \):
\[ 1 - p = p \implies 1 = 2p \implies p = \frac{1}{2} \]

Thus, the value of \( p \) is \( \frac{1}{2} \).

### Final Answer:
\[ p = \frac{1}{2} \]