If mean and variance of a Binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is

Question

Prepp Team · Accepted Answer

Understanding the Binomial Distribution Problem

The question provides the mean and variance of a Binomial variate X and asks for the probability that X takes a value greater than 1. A Binomial distribution is defined by two parameters: the number of trials (n) and the probability of success on a single trial (p).

For a Binomial variate X following $B(n, p)$, the mean and variance are given by:

Mean $E[X] = np$
Variance $Var[X] = np(1-p)$

Finding the Parameters n and p

We are given the mean and variance values:

Mean $np = 2$
Variance $np(1-p) = 1$

We can use these two equations to find the values of n and p. Substitute the first equation ($np = 2$) into the second equation:

$2(1-p) = 1$

Now, solve for p:

$2 - 2p = 1$

$1 = 2p$

$p = \frac{1}{2}$

Now that we have the value of p, substitute it back into the equation $np = 2$ to find n:

$n \times \frac{1}{2} = 2$

$n = 4$

So, the Binomial variate X follows a Binomial distribution with parameters $n = 4$ and $p = \frac{1}{2}$. The probability of failure is $1-p = 1 - \frac{1}{2} = \frac{1}{2}$.

Calculating the Probability $P(X > 1)$

We need to find the probability that X takes a value greater than 1, which is $P(X > 1)$. For a discrete distribution like the Binomial, this means $P(X=2) + P(X=3) + P(X=4)$. Alternatively, it is often easier to calculate the complement probability: $P(X > 1) = 1 - P(X \le 1)$.

$P(X \le 1) = P(X=0) + P(X=1)$

The probability mass function (PMF) for a Binomial distribution $B(n, p)$ is given by:

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

For our distribution $B(4, \frac{1}{2})$, the PMF is:

$P(X=k) = \binom{4}{k} (\frac{1}{2})^k (\frac{1}{2})^{4-k} = \binom{4}{k} (\frac{1}{2})^{4} = \binom{4}{k} \frac{1}{16}$

Now, let's calculate $P(X=0)$ and $P(X=1)$:

For $k=0$: $P(X=0) = \binom{4}{0} \frac{1}{16} = 1 \times \frac{1}{16} = \frac{1}{16}$
For $k=1$: $P(X=1) = \binom{4}{1} \frac{1}{16} = 4 \times \frac{1}{16} = \frac{4}{16}$

Now, calculate $P(X \le 1)$:

$P(X \le 1) = P(X=0) + P(X=1) = \frac{1}{16} + \frac{4}{16} = \frac{5}{16}$

Finally, calculate $P(X > 1)$:

$P(X > 1) = 1 - P(X \le 1) = 1 - \frac{5}{16} = \frac{16}{16} - \frac{5}{16} = \frac{11}{16}$

Result

The probability that the Binomial variate X takes a value greater than 1 is $\frac{11}{16}$.

Revision Table: Binomial Distribution

Concept	Formula/Description
Definition	Discrete probability distribution for the number of successes in a fixed number of independent Bernoulli trials.
Parameters	n (number of trials), p (probability of success)
Mean $E[X]$	$np$
Variance $Var[X]$	$np(1-p)$
PMF $P(X=k)$	$\binom{n}{k} p^k (1-p)^{n-k}$ for $k = 0, 1, \dots, n$

Additional Information: Properties and Applications

The Binomial distribution is widely used in various fields to model situations where there are a fixed number of independent trials, each with only two possible outcomes (success or failure), and the probability of success is constant for every trial. Examples include the number of heads in a series of coin flips, the number of defective items in a batch, or the number of correct answers on a multiple-choice test by guessing.

Understanding the relationship between the mean, variance, and the parameters n and p is crucial for solving problems like this one. The fact that the variance is always less than or equal to the mean ($np(1-p) \le np$ since $1-p \le 1$) is a characteristic property of the Binomial distribution, provided $p > 0$. In our case, $1 < 2$, which fits this property.