Question

Let \(X\) have a binomial distribution \(B(6,p)\). If the sum of the mean and the variance of \(X\) is \(\dfrac{21}{8}\), then \[ \frac{P(2\le X<4)}{P(4<X<6)} \] is equal to:

• A. \(65\)
• B. \(195\)
• C. \(\dfrac{195}{2}\)
• D. \(\dfrac{225}{2}\)

Hint:

For binomial problems, first determine \(p\) using mean and variance, then compute only the required terms instead of full probability tables.

Answer 1

The Correct Option is C

Step 1: Use mean and variance of binomial distribution For \(X\sim B(6,p)\): \[ \text{Mean}=6p,\qquad \text{Variance}=6p(1-p) \] Given: \[ 6p+6p(1-p)=\frac{21}{8} \] \[ 6p(2-p)=\frac{21}{8} \] \[ 48p(2-p)=21 \] \[ 96p-48p^2-21=0 \] \[ 48p^2-96p+21=0 \] Solving: \[ p=\frac{96\pm\sqrt{96^2-4\cdot48\cdot21}}{96} =\frac{96\pm72}{96} \] Rejecting \(p>1\), we get: \[ p=\frac14 \] Step 2: Compute probabilities \[ P(2\le X<4)=P(X=2)+P(X=3) \] \[ = \binom62\left(\frac14\right)^2\left(\frac34\right)^4 +\binom63\left(\frac14\right)^3\left(\frac34\right)^3 \] \[ =15\cdot\frac1{16}\cdot\frac{81}{256} +20\cdot\frac1{64}\cdot\frac{27}{64} =\frac{1215}{4096} \] \[ P(4<X<6)=P(X=5) =\binom65\left(\frac14\right)^5\left(\frac34\right) =\frac{18}{4096} \] Step 3: Required ratio \[ \frac{P(2\le X<4)}{P(4<X<6)} =\frac{1215}{18} =\frac{195}{2} \] \[ \boxed{\dfrac{195}{2}} \]