Question

In a Binomial distribution, if \(p=q\) and \(n\geq 4\), then \[ 2^nP(X=5)= \]

• A. \(5\)
• B. \({}^{n}C_{2}\)
• C. \(10\)
• D. \({}^{n}C_{5}\)

Hint:

Whenever \(p=q=\frac12\), the binomial probability simplifies to \[ P(X=r)=\frac{{}^{n}C_r}{2^n}. \]

Answer 1

The Correct Option is D

Step 1: Use the condition \(p=q\).
Since \[ p+q=1 \] and \[ p=q, \] we get \[ p=q=\frac12. \]

Step 2: Write the binomial probability.
For a binomial distribution, \[ P(X=r) = {}^{n}C_{r}p^{r}q^{\,n-r}. \] Therefore, \[ P(X=5) = {}^{n}C_{5} \left(\frac12\right)^5 \left(\frac12\right)^{n-5}. \] \[ P(X=5) = {}^{n}C_{5} \left(\frac12\right)^n. \]

Step 3: Multiply by \(2^n\).
\[ 2^nP(X=5) = 2^n\cdot {}^{n}C_{5} \left(\frac12\right)^n. \] \[ = {}^{n}C_{5}. \]

Step 4: Final conclusion.
Hence, \[ \boxed{{}^{n}C_{5}} \]