Question

If the mean and the variance of a binomial variate $X$ are $2$ and $1$ respectively, then the probability that $X$ takes a value greater than or equal to one is :

• A. $\frac{1}{16}$
• B. $\frac{9}{16}$
• C. $\frac{3}{4}$
• D. $\frac{15}{16}$

Answer 1

The Correct Option is D

We need to solve this problem step-by-step by using the properties of a binomial distribution. Let's begin:

A binomial distribution is defined by two parameters: the number of trials \(n\) and the probability of success \(p\) in each trial. The mean and the variance of a binomial distribution are given by the following formulas:

1. Mean: \(\mu = np\)
2. Variance: \(\sigma^2 = np(1-p)\)

For the given problem, the mean is \(2\) and the variance is \(1\). Let's use these equations:

Equation for mean: \(np = 2\) ...(1)

Equation for variance: \(np(1-p) = 1\) ...(2)

Substitute \(np = 2\) from equation (1) into equation (2):

\(2(1-p) = 1\)

Simplifying the equation: \(2 - 2p = 1\)

\(2p = 1\)

\(p = \frac{1}{2}\)

Now, substitute \(p = \frac{1}{2}\) back into equation (1):

\(n \cdot \frac{1}{2} = 2\)

\(n = 4\)

Now we know the parameters of the binomial distribution: \(n = 4\) and \(p = \frac{1}{2}\).

We want to find the probability that \(X \geq 1\). This is equal to \(1 - P(X = 0)\).

The probability of \(X = 0\) using the binomial probability formula is:

\(P(X = 0) = \binom{4}{0} \cdot \left(\frac{1}{2}\right)^0 \cdot \left(\frac{1}{2}\right)^4 = 1 \cdot 1 \cdot \frac{1}{16} = \frac{1}{16}\)

Therefore:

\(P(X \geq 1) = 1 - P(X = 0) = 1 - \frac{1}{16} = \frac{15}{16}\)

Thus, the probability that \(X\) takes a value greater than or equal to one is \(\frac{15}{16}\).