Question

In a Binomial distribution, if \(n\) is the number of trials and the mean and variance are \(4\) and \(3\) respectively, then \[ 2^{32}\cdot P\left(X=\frac{n}{2}\right)= \]

• A. \({}^{16}C_8(3^8)\)
• B. \({}^{12}C_6(2^6)\)
• C. \({}^{32}C_{16}(3^{16})\)
• D. \({}^{16}C_7(3^9)\)

Hint:

For Binomial distribution: \[ \text{Mean}=np, \qquad \text{Variance}=npq. \] Use these first to determine \(n\), \(p\), and \(q\), then apply the probability formula: \[ P(X=r)={}^{n}C_r p^r q^{n-r}. \]

Answer 1

The Correct Option is A

Step 1: Use the mean and variance formulas for Binomial distribution.
For a Binomial distribution, \[ \text{Mean}=np \] and \[ \text{Variance}=npq \] where \[ q=1-p \] Given, \[ np=4 \] and \[ npq=3 \]

Step 2: Find \(q\) and \(p\).
Using \[ npq=3 \] and \[ np=4, \] we get \[ 4q=3 \] \[ q=\frac34 \] Therefore, \[ p=1-\frac34=\frac14 \]

Step 3: Find the number of trials \(n\).
Since \[ np=4, \] we have \[ n\left(\frac14\right)=4 \] \[ n=16 \] Thus, \[ \frac n2=8 \]

Step 4: Compute \(P(X=8)\).
For binomial distribution, \[ P(X=r)={}^{n}C_r p^r q^{n-r} \] Hence, \[ P(X=8) = {}^{16}C_8 \left(\frac14\right)^8 \left(\frac34\right)^8 \] \[ = {}^{16}C_8 \frac{1}{4^8} \frac{3^8}{4^8} \] \[ = {}^{16}C_8 \frac{3^8}{4^{16}} \]

Step 5: Multiply by \(2^{32}\).
Since \[ 4^{16}=(2^2)^{16}=2^{32}, \] we get \[ 2^{32}P(X=8) = 2^{32}\cdot {}^{16}C_8 \frac{3^8}{2^{32}} \] \[ = {}^{16}C_8(3^8) \]

Step 6: Final conclusion.
Therefore, \[ \boxed{{}^{16}C_8(3^8)} \]