1. Key Concepts and Formulas

A binomial distribution, denoted as $B(n,p)$, models the number of successes in $n$ independent Bernoulli trials, where each trial has a probability $p$ of success. The key parameters and their formulas are:

• Number of Trials ($n$): A positive integer representing the total number of independent trials ($n \in \mathbb{Z}^+$).
• Probability of Success ($p$): The probability of success in a single trial, where $0 \le p \le 1$.
• Probability of Failure ($q$): The probability of failure in a single trial, related to $p$ by the equation $q = 1-p$. Consequently, $0 \le q \le 1$.
• Mean ($\mu$): The expected number of successes, given by the formula: $\mu = np$
• Variance ($\sigma^2$): A measure of the spread of the distribution, given by the formula: $\sigma^2 = npq$

2. Step-by-Step Solution

Step 1: Translate the given information into algebraic equations.

• What we are doing: We are converting the problem's descriptive statements about the mean and variance into precise mathematical equations.
• Why we are doing it: This is the foundational step to set up the system of equations that we will solve.

The problem states:

1. The sum of the mean and the variance is 5. $\mu + \sigma^2 = 5$ Substituting the formulas for $\mu$ and $\sigma^2$: $np + npq = 5 \quad \text{(Equation 1)}$
2. The product of the mean and the variance is 6. $\mu \cdot \sigma^2 = 6$ Substituting the formulas for $\mu$ and $\sigma^2$: $(np)(npq) = 6 \quad \text{(Equation 2)}$

Step 2: Simplify the initial equations.

• What we are doing: We are simplifying Equation 1 by factoring and Equation 2 by combining terms, to make them easier to work with.

• Why we are doing it: Simplification often reveals relationships between variables and prepares the equations for strategic manipulation.

• Simplifying Equation 1: Factor out the common term $np$: $np(1+q) = 5 \quad \text{(Equation 1')}$

• Simplifying Equation 2: Multiply the terms: $n^2p^2q = 6 \quad \text{(Equation 2')}$

Step 3: Strategically eliminate variables to solve for $q$.

• What we are doing: We have two equations (1' and 2') with three unknowns ($n, p, q$). Our goal is to reduce the number of variables to solve for one. We will use division to eliminate $n$ and $p$.
• Why we are doing it: Notice that Equation 1' contains $np$ and Equation 2' contains $n^2p^2$. If we square Equation 1', we will get an $n^2p^2$ term, allowing us to divide Equation 3 by Equation 2' to cancel out $n^2p^2$.

1. Square Equation 1': $(np(1+q))^2 = 5^2$ $n^2p^2(1+q)^2 = 25 \quad \text{(Equation 3)}$
2. Divide Equation 3 by Equation 2': $\frac{n^2p^2(1+q)^2}{n^2p^2q} = \frac{25}{6}$ The $n^2p^2$ terms cancel out: $\frac{(1+q)^2}{q} = \frac{25}{6}$
3. Form a quadratic equation for $q$: Cross-multiply: $6(1+q)^2 = 25q$ Expand $(1+q)^2$: $6(1 + 2q + q^2) = 25q$ Distribute the 6: $6 + 12q + 6q^2 = 25q$ Rearrange into standard quadratic form ($aq^2+bq+c=0$): $6q^2 + 12q - 25q + 6 = 0$ $6q^2 - 13q + 6 = 0$

Step 4: Solve for $q$ and validate the solution.

• What we are doing: We solve the quadratic equation obtained in Step 3 for $q$.
• Why we are doing it: This will give us the possible values for $q$. It is crucial to validate these values against the definition of a probability.

We solve the quadratic equation $6q^2 - 13q + 6 = 0$ using factorization: We need two numbers that multiply to $6 \times 6 = 36$ and add up to $-13$. These numbers are $-9$ and $-4$. $6q^2 - 9q - 4q + 6 = 0$ Factor by grouping: $3q(2q-3) - 2(2q-3) = 0$ $(3q-2)(2q-3) = 0$ This yields two possible values for $q$: $3q-2=0 \implies q = \frac{2}{3}$ $2q-3=0 \implies q = \frac{3}{2}$

Validation: Since $q$ represents a probability, it must satisfy $0 \le q \le 1$.

• $q = \frac{2}{3}$ (approximately 0.667) is within the valid range.
• $q = \frac{3}{2}$ (1.5) is greater than 1, so it is an invalid probability and must be rejected.

Therefore, the only valid value for $q$ is $q = \frac{2}{3}$.

Step 5: Determine the remaining parameters $p$ and $n$.

• What we are doing: Using the valid value of $q$, we find $p$ and then $n$.
• Why we are doing it: We need all parameters ($n, p, q$) to evaluate the final expression.

1. Find $p$: Using the relationship $p = 1-q$: $p = 1 - \frac{2}{3}$ $p = \frac{1}{3}$
2. Find $n$: Substitute the values of $p$ and $q$ into Equation 1': $np(1+q) = 5$. $n \left(\frac{1}{3}\right) \left(1 + \frac{2}{3}\right) = 5$ $n \left(\frac{1}{3}\right) \left(\frac{5}{3}\right) = 5$ $\frac{5n}{9} = 5$ Multiply both sides by $\frac{9}{5}$: $n = 5 \times \frac{9}{5}$ $n = 9$

Validation: Since $n$ represents the number of trials, it must be a positive integer. $n=9$ is a valid value.

Step 6: Calculate the final expression $6(n+p-q)$.

• What we are doing: We substitute the determined values of $n, p, q$ into the expression required by the question.
• Why we are doing it: This is the final step to answer the problem.

We have $n=9$, $p=\frac{1}{3}$, and $q=\frac{2}{3}$. Substitute these into $6(n+p-q)$: $6 \left(9 + \frac{1}{3} - \frac{2}{3}\right)$ Simplify the terms inside the parenthesis: $9 + \frac{1}{3} - \frac{2}{3} = 9 - \left(\frac{2}{3} - \frac{1}{3}\right) = 9 - \frac{1}{3}$ Convert to a common denominator: $9 - \frac{1}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$ Now, multiply by 6: $6 \left(\frac{26}{3}\right) = (6 \div 3) \times 26 = 2 \times 26 = 52$

3. Common Mistakes & Tips

• Parameter Validation: Always remember that $p$ and $q$ must be probabilities ($0 \le p, q \le 1$), and $n$ must be a positive integer. Reject any values that violate these conditions.
• Algebraic Errors: Be careful with squaring equations and dividing them. Ensure all terms are handled correctly, especially when expanding binomials like $(1+q)^2$.
• Factoring: Factoring $np$ from the sum equation ($np+npq = np(1+q)$) is a crucial step that simplifies subsequent calculations.

4. Summary

This problem required a thorough understanding of the binomial distribution's mean and variance formulas. We first translated the given sum and product information into two algebraic equations involving $n, p,$ and $q$. By strategically manipulating these equations, specifically by squaring one and dividing by the other, we formed a quadratic equation for $q$. After solving for $q$, we validated the solutions to ensure they represented a valid probability. With $q$ determined, we easily found $p$ and then $n$. Finally, we substituted these derived parameter values into the required expression to arrive at the solution.

The final answer is $\boxed{\text{52}}$, which corresponds to option (A).