Key Concepts and Formulas

A random variable $X$ follows a binomial distribution, denoted as $B(n, p)$, if it represents the number of "successes" in a fixed number of $n$ independent Bernoulli trials. Each trial has a constant probability of success $p$, and consequently, a probability of failure $q = 1-p$.

For a binomial distribution $B(n, p)$, the following fundamental formulas are used:

1. Probability Mass Function (PMF): The probability of observing exactly $k$ successes in $n$ trials is given by: $P(X=k) = {^nC_k} \, p^k \, (1-p)^{n-k}$ where $k \in \{0, 1, 2, \dots, n\}$ and ${^nC_k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

2. Mean ($\mu$): The expected number of successes is calculated as: $\mu = np$

3. Variance ($\sigma^2$): The measure of the spread of the distribution is: $\sigma^2 = np(1-p)$

These formulas are essential for characterizing and calculating probabilities related to binomial random variables.

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Step-by-Step Solution

Our goal is to determine the value of $54 \, P(X \leq 2)$ for a given binomially distributed random variable $X$.

Step 1: Determine the Parameters of the Binomial Distribution ($n$ and $p$)

Before calculating probabilities, we must first identify the specific binomial distribution by finding its parameters: $n$ (the total number of trials) and $p$ (the probability of success).

We are given:

• Mean of $X$: $\mu = 4$
• Variance of $X$: $\sigma^2 = \frac{4}{3}$

1. Formulate equations from the given mean and variance: Using the standard formulas for the mean and variance of a binomial distribution, we set up a system of equations: (1) $np = 4$ (2) $np(1-p) = \frac{4}{3}$ Reasoning: These equations directly translate the problem's given information into a solvable system for our unknown parameters $n$ and $p$.

2. Solve for $p$ (probability of success): We can efficiently solve this system by substituting the expression for $np$ from equation (1) into equation (2): $4(1-p) = \frac{4}{3}$ Now, divide both sides by 4 to isolate $(1-p)$: $1-p = \frac{4}{3 \times 4}$ $1-p = \frac{1}{3}$ To find $p$, subtract $\frac{1}{3}$ from 1: $p = 1 - \frac{1}{3}$ $p = \frac{2}{3}$ Reasoning: Solving for $p$ first simplifies the subsequent calculation of $n$. The value $p = \frac{2}{3}$ tells us the probability of success in any single trial. Consequently, the probability of failure is $q = 1-p = 1-\frac{2}{3} = \frac{1}{3}$.

3. Solve for $n$ (number of trials): Substitute the calculated value of $p = \frac{2}{3}$ back into equation (1) ($np=4$): $n \times \frac{2}{3} = 4$ Multiply both sides by $\frac{3}{2}$ to solve for $n$: $n = 4 \times \frac{3}{2}$ $n = 6$ Reasoning: With $p$ determined, finding $n$ completes the identification of the specific binomial distribution. The value $n=6$ indicates that there are 6 independent trials in this experiment.

   Thus, the random variable $X$ follows a binomial distribution $X \sim B\left(6, \frac{2}{3}\right)$.

Step 2: Calculate $P(X \leq 2)$

The expression $P(X \leq 2)$ represents the cumulative probability that the number of successes is less than or equal to 2. For a discrete binomial distribution, this means summing the probabilities of obtaining exactly 0, 1, or 2 successes: $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$ Reasoning: The notation $P(X \leq k)$ requires summing individual probabilities for all outcomes from $0$ up to $k$.

We will use the PMF formula $P(X=k) = {^nC_k} \, p^k \, (1-p)^{n-k}$ with $n=6$, $p=\frac{2}{3}$, and $(1-p)=\frac{1}{3}$.

1. Calculate $P(X=0)$: $P(X=0) = {^6C_0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^{6-0}$ Since ${^6C_0} = 1$ and $(\frac{2}{3})^0 = 1$: $P(X=0) = 1 \times 1 \times \left(\frac{1}{3}\right)^6 = \frac{1^6}{3^6} = \frac{1}{729}$

2. Calculate $P(X=1)$: $P(X=1) = {^6C_1} \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^{6-1}$ Since ${^6C_1} = 6$: $P(X=1) = 6 \times \frac{2}{3} \times \left(\frac{1}{3}\right)^5$ $P(X=1) = 4 \times \frac{1}{3^5} = \frac{4}{243}$ To prepare for summation, we express this with a common denominator of $3^6 = 729$: $P(X=1) = \frac{4 \times 3}{243 \times 3} = \frac{12}{729}$

3. Calculate $P(X=2)$: $P(X=2) = {^6C_2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^{6-2}$ Since ${^6C_2} = \frac{6 \times 5}{2 \times 1} = 15$: $P(X=2) = 15 \times \left(\frac{4}{9}\right) \times \left(\frac{1}{3}\right)^4$ $P(X=2) = 15 \times \frac{4}{9} \times \frac{1}{81}$ $P(X=2) = \frac{60}{9 \times 81} = \frac{60}{729}$

4. Sum the probabilities to find $P(X \leq 2)$: $P(X \leq 2) = \frac{1}{729} + \frac{12}{729} + \frac{60}{729}$ $P(X \leq 2) = \frac{1 + 12 + 60}{729} = \frac{73}{729}$ Reasoning: This is the intermediate cumulative probability required before the final calculation.

Step 3: Calculate the Final Required Value

The problem asks for the value of $54 \, P(X \leq 2)$. We substitute the calculated value of $P(X \leq 2)$: $54 \, P(X \leq 2) = 54 \times \frac{73}{729}$ Reasoning: This is the final computation specified by the problem statement.

To simplify the expression, we can recognize common factors. Note that $729 = 3^6 = 27 \times 27$, and $54 = 2 \times 27$: $54 \, P(X \leq 2) = (2 \times 27) \times \frac{73}{(27 \times 27)}$ Cancel one factor of 27 from the numerator and the denominator: $54 \, P(X \leq 2) = \frac{2 \times 73}{27}$ $54 \, P(X \leq 2) = \frac{146}{27}$ Reasoning: Simplifying the fraction provides the final answer in its most reduced form, which matches the format of the options.

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Common Mistakes & Tips

• Parameter Identification Errors: A frequent mistake is incorrectly solving for $n$ and $p$. Always double-check your calculations when using the mean and variance formulas to ensure $n$ and $p$ are correct. Incorrect parameters will lead to incorrect probabilities.
• Arithmetic Precision: Binomial probability calculations involve combinations, powers of fractions, and sums. Be meticulous with each step to avoid numerical errors. Remember that ${^nC_0} = 1$, ${^nC_n} = 1$, and ${^nC_1} = n$.
• Misinterpreting Probability Notation: Understand the distinction between $P(X=k)$ (exact probability) and $P(X \leq k)$ (cumulative probability). For $P(X \leq k)$, always sum probabilities from $X=0$ up to $X=k$.
• Strategic Simplification: When summing multiple probabilities involving fractions, find a common denominator early (e.g., $3^6 = 729$) to simplify addition. For final multiplication, look for common factors between the numerator and denominator to reduce the fraction efficiently.

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Summary

This problem effectively tests the understanding and application of the binomial distribution. The solution followed a structured approach: first, we accurately determined the parameters of the binomial distribution, $n=6$ and $p=\frac{2}{3}$, by utilizing the given mean and variance. Next, we calculated the cumulative probability $P(X \leq 2)$ by individually computing $P(X=0)$, $P(X=1)$, and $P(X=2)$ using the Probability Mass Function and summing them. Finally, we multiplied this cumulative probability by 54 and simplified the resulting fraction to arrive at the final answer. The ability to correctly identify distribution parameters and perform careful probability calculations is key to solving such problems.

The final answer is $\boxed{\frac{146}{27}}$, which corresponds to option (A).