Animated Solution for Mathematics - Probability: Let $X$ be a binomially distributed random variable with mean 4 and variance $\frac{4}{3}$. Then $54P(X\le 2)$ is equal to

Mastering the Binomial Distribution

My dear student, welcome to the fascinating world of discrete probability. Today, we are going to dissect a problem that might seem like a dry calculation, but is actually a beautiful exercise in logical deduction.

We are dealing with a Binomial Distribution, the mathematical heartbeat of experiments where we have a fixed number of trials, each with a binary outcome: success or failure. Our goal is to find the value of $54P(X\le 2)$ given the mean and variance.

Let us peel back the layers of this problem together.

We are given two vital clues: the mean $E(X)=np=4$ and the variance $Var(X)=npq=\frac{4}{3}$. Here, $n$ is the number of trials, $p$ is the probability of success, and $q$ is the probability of failure.

The beauty of this setup lies in the relationship between these two parameters. If we divide the variance by the mean, we get:

Since the sum of probabilities must be unity, we know $p=1-q=1-\frac{1}{3}=\frac{2}{3}$.

Now, with $p$ in hand, we return to our mean equation: $n\times \frac{2}{3}=4$. Solving for $n$, we find:

We have successfully defined our experiment: $6$ trials, with a success probability of $\frac{2}{3}$ and a failure probability of $\frac{1}{3}$.

The question asks for $54P(X\le 2)$. This notation, $P(X\le 2)$, is simply a request to sum the probabilities of the individual, mutually exclusive events where the number of successes is $0$, $1$, or $2$.

Mathematically, this is $P(X=0)+P(X=1)+P(X=2)$. We will use the general binomial probability formula:

Let us calculate these one by one with precision.

For $k=0$:

For $k=1$:

For $k=2$:

Notice how the denominator $729$ remains consistent? This is a gift of the binomial expansion. Summing these, we get:

We are almost at the finish line. The problem asks for $54P(X\le 2)$. Substituting our sum, we have:

Here is where we appreciate the elegance of numbers. We can write $54$ as $2\times 27$ and $729$ as $27\times 27$. The expression becomes:

And there it is! The final result is $\frac{146}{27}$.

I hope you can see that this wasn't just a calculation; it was a journey of uncovering hidden parameters to reveal a simple, elegant truth. Keep practicing, and keep falling in love with the process.