1. Key Concepts and Formulas

This problem is based on the Binomial Distribution, which describes the number of successes in a fixed number of independent Bernoulli trials.

• Binomial Probability Mass Function (PMF): For a random variable $X \sim B(n, p)$, the probability of getting exactly $k$ successes in $n$ trials is given by: $P(X=k) = \binom{n}{k} p^k q^{n-k}$ where $n$ is the number of trials, $p$ is the probability of success in a single trial, $q = 1-p$ is the probability of failure, and $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

• Mean (Expected Value) of X: The average number of successes expected: $E(X) = np$

• Variance of X: A measure of the spread of the distribution: $Var(X) = npq$

2. Step-by-Step Solution

Step 1: Identify Given Information and Parameters

The problem states that $X$ is a random variable following a binomial distribution $B(7, p)$.

• This immediately tells us the number of trials, $n = 7$.
• We are also given a relationship between probabilities: $P(X=3) = 5P(X=4)$. Our primary goal is to use this relationship to determine the unknown probability $p$.

Step 2: Apply the Probability Mass Function (PMF)

We use the PMF formula, $P(X=k) = \binom{n}{k} p^k q^{n-k}$, for $k=3$ and $k=4$, with $n=7$.

• For $P(X=3)$: Here, $k=3$. $P(X=3) = \binom{7}{3} p^3 q^{7-3} = \binom{7}{3} p^3 q^4$ Calculate the binomial coefficient $\binom{7}{3}$: $\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ So, $P(X=3) = 35 p^3 q^4$.

• For $P(X=4)$: Here, $k=4$. $P(X=4) = \binom{7}{4} p^4 q^{7-4} = \binom{7}{4} p^4 q^3$ Calculate the binomial coefficient $\binom{7}{4}$: $\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ (Alternatively, using the property $\binom{n}{k} = \binom{n}{n-k}$, we have $\binom{7}{4} = \binom{7}{7-4} = \binom{7}{3} = 35$.) So, $P(X=4) = 35 p^4 q^3$.

Step 3: Set Up and Solve the Probability Equation

Now, substitute the expressions for $P(X=3)$ and $P(X=4)$ into the given equation: $P(X=3) = 5P(X=4)$ $35 p^3 q^4 = 5 \cdot (35 p^4 q^3)$

To simplify, divide both sides by $35$: $p^3 q^4 = 5 p^4 q^3$

Assuming $p \neq 0$ and $q \neq 0$ (since probabilities of specific outcomes are non-zero), we can divide both sides by $p^3 q^3$: $\frac{p^3 q^4}{p^3 q^3} = \frac{5 p^4 q^3}{p^3 q^3}$ $q = 5p$

Step 4: Determine the Values of $p$ and $q$

We have the relationship $q = 5p$. We also know the fundamental property that the probability of success and failure must sum to 1: $p + q = 1$

Substitute $q = 5p$ into this equation: $p + 5p = 1$ $6p = 1$ $p = \frac{1}{6}$

Now, find $q$: $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$

So, the parameters of our binomial distribution are $n=7$, $p=\frac{1}{6}$, and $q=\frac{5}{6}$.

Step 5: Calculate the Mean and Variance of X

Using the formulas for the mean and variance:

• Mean ($E(X)$): $E(X) = np = 7 \cdot \frac{1}{6} = \frac{7}{6}$

• Variance ($Var(X)$): $Var(X) = npq = 7 \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{35}{36}$

Step 6: Calculate the Sum of the Mean and Variance

The problem asks for the sum of the mean and the variance of $X$: $\text{Sum} = E(X) + Var(X) = \frac{7}{6} + \frac{35}{36}$

To add these fractions, find a common denominator, which is 36. Multiply the numerator and denominator of the first fraction by 6: $\text{Sum} = \frac{7 \cdot 6}{6 \cdot 6} + \frac{35}{36}$ $\text{Sum} = \frac{42}{36} + \frac{35}{36}$ $\text{Sum} = \frac{42 + 35}{36}$ $\text{Sum} = \frac{77}{36}$

3. Common Mistakes & Tips

• Incorrectly calculating binomial coefficients: Ensure $n! / (k!(n-k)!)$ is calculated correctly. Remember $\binom{n}{k} = \binom{n}{n-k}$ for symmetry.
• Algebraic errors: Be careful when simplifying equations involving powers of $p$ and $q$.
• Forgetting $p+q=1$: This fundamental identity is crucial for solving for $p$ and $q$.
• Confusing mean and variance formulas: Always use $np$ for mean and $npq$ for variance.

4. Summary

This problem required us to utilize the core definitions and formulas of the binomial distribution. By first identifying the parameters $n$ and the probability mass function, we translated the given probabilistic relationship into an algebraic equation. Solving this equation allowed us to find the unknown probability of success, $p$. Once $p$ and $q$ were determined, we could easily calculate the mean and variance of the distribution and sum them to obtain the final answer. The key was careful application of the PMF and algebraic simplification.

5. Final Answer

The sum of the mean and variance of $X$ is $\frac{77}{36}$. Comparing this result with the given options, we find that $\frac{77}{36}$ corresponds to option (C).

The final answer is \boxed{77 \over 36}.