Question

If the mean and variance of a binomial variate \(X\) are \(2\) and \(1\) respectively, then the probability that \(X\) takes a value greater than one is

• A. \(\dfrac{1}{16}\)
• B. \(\dfrac{11}{16}\)
• C. \(\dfrac{5}{16}\)
• D. \(\dfrac{15}{16}\)

Hint:

For a binomial distribution, \[ \boxed{ \text{Mean}=np,\qquad \text{Variance}=np(1-p). } \] If \(np\) and the variance are known, first determine \(p\), then \(n\).

Answer 1

The Correct Option is B

Step 1: Find the parameters of the binomial distribution. For a binomial random variable, \[ \text{Mean}=np=2, \] and \[ \text{Variance}=np(1-p)=1. \] Using \[ np=2, \] we get \[ 2(1-p)=1 \] which gives \[ p=\frac12. \] Hence, \[ n=\frac{2}{p} =\frac{2}{1/2} =4. \] Thus, \[ X\sim B\!\left(4,\frac12\right). \]

Step 2: Compute \(P(X>1)\). \[ P(X>1) = 1-\left[P(X=0)+P(X=1)\right]. \] Now, \[ P(X=0) = \binom40 \left(\frac12\right)^4 = \frac1{16}, \] and \[ P(X=1) = \binom41 \left(\frac12\right)^4 = \frac4{16}. \] Therefore, \[ P(X>1) = 1-\frac{1+4}{16} = 1-\frac5{16} = \frac{11}{16}. \] Hence, \[ \boxed{\frac{11}{16}}. \] Therefore, \[ \boxed{(B)} \] is the correct answer.