Question

For a Binomial distribution, mean is \(15\) and variance is \(6\). If \[ P(X\ge2) = 1-\left(\frac25\right)^{25}k, \] then \(k=\)

• A. \(2^{24}\)
• B. \(\dfrac{77}{2}\)
• C. \(\left(\dfrac25\right)^{24}\)
• D. \(\dfrac{78}{5}\)

Hint:

For Binomial distribution: \[ np=\text{Mean}, \qquad npq=\text{Variance}. \] To calculate \(P(X\ge2)\), it is often easier to use the complement: \[ P(X\ge2)=1-[P(X=0)+P(X=1)]. \]

Answer 1

The Correct Option is B

Concept: For a binomial distribution, \[ \mu=np, \] \[ \sigma^2=npq. \] Also, \[ P(X\ge2) = 1-[P(X=0)+P(X=1)]. \]

Step 1: Determine \(n\) and \(p\).
Given \[ np=15 \] and \[ npq=6. \] Therefore, \[ q=\frac6{15}=\frac25. \] Hence \[ p=\frac35. \] Now \[ n=\frac{15}{3/5}=25. \]

Step 2: Find \(P(X=0)\).
\[ P(X=0) = q^{25} = \left(\frac25\right)^{25}. \]

Step 3: Find \(P(X=1)\).
\[ P(X=1) = {25 \choose 1} \left(\frac35\right) \left(\frac25\right)^{24}. \] \[ = 25\cdot\frac35 \left(\frac25\right)^{24}. \] \[ = 15 \left(\frac25\right)^{24}. \]

Step 4: Compute \(P(X\ge2)\).
\[ P(X\ge2) = 1-\left[ \left(\frac25\right)^{25} + 15\left(\frac25\right)^{24} \right]. \] Factor out \[ \left(\frac25\right)^{24}. \] \[ = 1- \left(\frac25\right)^{24} \left( \frac25+15 \right). \] \[ = 1- \left(\frac25\right)^{24} \left( \frac{77}{5} \right). \] \[ = 1- \left(\frac25\right)^{25} \left( \frac{77}{2} \right). \] Comparing with \[ 1-\left(\frac25\right)^{25}k, \] we obtain \[ \boxed{k=\frac{77}{2}}. \]