If X is a binomial variate with n = 7 and `P(X=3)=P(X=4)` then `P(X=5)=`

Options

A. `21.(3^(2))/(4^(7))`
B. `21.(3^(5))/(4^(7))`
C. `7.(2^(5))/(3^(6))`
D. `(21)/(2^(7))`

Answer

To solve the problem, we need to find \( P(X = 5) \) given that \( P(X = 3) = P(X = 4) \) for a binomial random variable \( X \) with \( n = 7 \).

### Step-by-Step Solution:

1. **Understand the Binomial Probability Formula**:
The probability mass function for a binomial random variable is given by:
\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]
where \( n \) is the number of trials, \( r \) is the number of successes, \( p \) is the probability of success, and \( \binom{n}{r} \) is the binomial coefficient.

2. **Set Up the Equation**:
Given \( P(X = 3) = P(X = 4) \), we can write:
\[ \binom{7}{3} p^3 (1-p)^{4} = \binom{7}{4} p^4 (1-p)^{3} \]

3. **Calculate the Binomial Coefficients**:
The binomial coefficients are:
\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]
\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 35 \]
Therefore, \( \binom{7}{3} = \binom{7}{4} \).

4. **Simplify the Equation**:
Since both coefficients are equal, we can cancel them out:
\[ p^3 (1-p)^{4} = p^4 (1-p)^{3} \]

5. **Rearranging the Equation**:
Dividing both sides by \( p^3 \) (assuming \( p \neq 0 \)):
\[ (1-p)^{4} = p (1-p)^{3} \]
Dividing both sides by \( (1-p)^{3} \) (assuming \( 1-p \neq 0 \)):
\[ 1-p = p \]

6. **Solve for \( p \)**:
Rearranging gives:
\[ 1 = 2p \implies p = \frac{1}{2} \]

7. **Calculate \( P(X = 5) \)**:
Now that we have \( p = \frac{1}{2} \), we can find \( P(X = 5) \):
\[ P(X = 5) = \binom{7}{5} p^5 (1-p)^{2} \]
\[ = \binom{7}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{2} \]
\[ = \binom{7}{5} \left(\frac{1}{2}\right)^7 \]

8. **Calculate the Binomial Coefficient**:
\[ \binom{7}{5} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \]

9. **Final Calculation**:
\[ P(X = 5) = 21 \cdot \left(\frac{1}{2}\right)^7 = \frac{21}{128} \]

### Final Answer:
\[ P(X = 5) = \frac{21}{128} \]