Animated Solution for Mathematics - Probability: 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 one is equal to .........

Welcome, future engineer! Today, we are going to explore a problem that perfectly captures the elegance of probability. We are dealing with a binomial variate $X$, a fundamental concept in statistics that models the number of successes in a fixed number of independent trials.

Imagine you are conducting an experiment with $n$ trials, where each trial has a probability of success $p$ and a probability of failure $q=1-p$. The mean of this distribution is $\mu =np$, and the variance is ${\sigma }^2=npq$.

We are given $\mu =2$ and ${\sigma }^2=1$. Our mission is to find the probability that $X$ is strictly greater than one.

We have two powerful equations at our disposal: $np=2$ and $npq=1$. Instead of struggling with these, let us perform a beautiful algebraic maneuver.

If we divide the variance by the mean, we get:

Notice how the $n$ and $p$ terms vanish into thin air, leaving us with $q=1/2$. Since $p+q=1$, we immediately find that $p=1-1/2=1/2$.

Now that we have $p=1/2$, we return to our mean equation: $n\cdot (1/2)=2$. Solving this, we find $n=4$. We now know everything about our binomial distribution: it consists of four trials, and each trial is a fair coin flip with a success probability of $1/2$.

Now, the question asks for $P(X>1)$. Since $X$ can take values from $0$ to $4$, calculating $P(X=2)+P(X=3)+P(X=4)$ is a valid approach.

But wait! Why do extra work when we can use the complement rule? The sum of all probabilities in a distribution is always $1$.

Therefore, $P(X>1)=1-P(X\le 1)$. This simplifies our task to calculating only $P(X=0)$ and $P(X=1)$. This is our strategic pivot—a way to minimize effort and maximize accuracy.

Using the binomial probability mass function $P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}$, we calculate the components.

For $k=0$:

For $k=1$:

Now, we combine these: $P(X\le 1)=1/16+4/16=5/16$. Finally, the probability that $X$ is greater than one is:

We have arrived at our answer: $11/16$. By breaking the problem into parameters, using algebraic shortcuts, and employing the complement rule, we turned a potentially tedious calculation into a smooth, logical flow.

Always remember, in physics and mathematics, the most elegant path is often the one that reveals the underlying structure of the problem. Keep practicing, stay curious, and keep falling in love with the logic behind the numbers!