Question

If the mean and variance of a variable following binomial distribution are 16 and 8 respectively, then find $P(X > 0)$.

• A. $(1/2)^{32}$
• B. $1 - (1/2)^{32}$
• C. $(1/2)^{16}$
• D. $1 - (1/2)^{16}$

Hint:

To quickly find $n$ and $p$ in a Binomial distribution, use $q = \text{Variance}/\text{Mean}$. Once you have $q$ and $n$, the probability $P(X=0)$ is simply $q^n$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The parameters of a Binomial distribution, $n$ (number of trials) and $p$ (probability of success), can be determined using its mean and variance.
Key Formula or Approach:
For a Binomial distribution: \[ \text{Mean} = np \] \[ \text{Variance} = npq \] where $q = 1-p$ is the probability of failure.

Step 2: Detailed Explanation:
Given: \[ np = 16 \] \[ npq = 8 \]
Dividing the variance by the mean: \[ q = \frac{npq}{np} = \frac{8}{16} = \frac{1}{2} \] Since $p = 1 - q$: \[ p = 1 - \frac{1}{2} = \frac{1}{2} \]
Using $np = 16$ to find $n$: \[ n \left(\frac{1}{2}\right) = 16 \implies n = 32 \]
We need to find $P(X > 0)$: \[ P(X > 0) = 1 - P(X = 0) \] Using the Binomial probability mass function for $X=0$: \[ P(X = 0) = \binom{32}{0} p^0 q^{32} = q^{32} = \left(\frac{1}{2}\right)^{32} \]
Substituting this back: \[ P(X > 0) = 1 - \left(\frac{1}{2}\right)^{32} \]

Step 3: Final Answer:
The probability is $1 - (1/2)^{32}$.