1. Key Concepts and Formulas

• Binomial Distribution ($X \sim B(n, p)$): Represents the number of successes ($X$) in $n$ independent Bernoulli trials, where $p$ is the probability of success in a single trial and $q = 1-p$ is the probability of failure.
• Mean ($\mu$): For a binomial distribution, $\mu = E(X) = np$.
• Variance ($\sigma^2$): For a binomial distribution, $\sigma^2 = Var(X) = npq$.
• Probability Mass Function (PMF): The probability of getting exactly $k$ successes in $n$ trials is $P(X=k) = \binom{n}{k} p^k q^{n-k}$, for $k = 0, 1, 2, \ldots, n$.

2. Step-by-Step Solution

Step 1: Formulate equations for mean and variance. We are given the sum and product of the mean ($\mu$) and variance ($\sigma^2$) of a binomial distribution.

1. Sum: $\mu + \sigma^2 = 24$
2. Product: $\mu \cdot \sigma^2 = 128$

Step 2: Solve for the mean ($\mu$) and variance ($\sigma^2$). Let $\mu$ and $\sigma^2$ be the roots of a quadratic equation $t^2 - (\mu + \sigma^2)t + (\mu \cdot \sigma^2) = 0$. Substituting the given values: $t^2 - 24t + 128 = 0$ To solve this quadratic equation, we can factor it. We look for two numbers that multiply to 128 and add up to -24. These numbers are -16 and -8. $(t - 16)(t - 8) = 0$ This gives two possible values for $t$: $t = 16$ or $t = 8$. Thus, the possible pairs for $(\mu, \sigma^2)$ are $(16, 8)$ or $(8, 16)$.

Step 3: Determine the parameters $n$ and $p$ of the binomial distribution. We use the relationships $\mu = np$ and $\sigma^2 = npq$. A useful relation derived from these is $q = \frac{\sigma^2}{\mu}$.

Case 1: $\mu = 16$ and $\sigma^2 = 8$ First, calculate $q$: $q = \frac{\sigma^2}{\mu} = \frac{8}{16} = \frac{1}{2}$ Since $p = 1 - q$: $p = 1 - \frac{1}{2} = \frac{1}{2}$ Now, use $\mu = np$ to find $n$: $16 = n \left(\frac{1}{2}\right)$ $n = 16 \times 2 = 32$ This set of parameters ($n=32$, $p=1/2$) is valid because $n$ is a positive integer and $0 < p < 1$.

Case 2: $\mu = 8$ and $\sigma^2 = 16$ Calculate $q$: $q = \frac{\sigma^2}{\mu} = \frac{16}{8} = 2$ This value of $q$ is not possible for a probability, as $q$ must be between 0 and 1. Therefore, this case is invalid.

So, the unique parameters for the binomial distribution are $n=32$ and $p=1/2$. Consequently, $q = 1 - p = 1/2$.

Step 4: Calculate the probability of one or two successes. We need to find $P(X=1 \text{ or } X=2)$, which for mutually exclusive events is $P(X=1) + P(X=2)$. Using the binomial probability formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$ with $n=32$, $p=1/2$, and $q=1/2$: $P(X=k) = \binom{32}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{32-k} = \binom{32}{k} \left(\frac{1}{2}\right)^{32} = \frac{\binom{32}{k}}{2^{32}}$

For $P(X=1)$: $P(X=1) = \frac{\binom{32}{1}}{2^{32}} = \frac{32}{2^{32}}$

For $P(X=2)$: $P(X=2) = \frac{\binom{32}{2}}{2^{32}} = \frac{\frac{32 \times 31}{2 \times 1}}{2^{32}} = \frac{16 \times 31}{2^{32}} = \frac{496}{2^{32}}$

Now, sum these probabilities: $P(X=1) + P(X=2) = \frac{32}{2^{32}} + \frac{496}{2^{32}} = \frac{32 + 496}{2^{32}} = \frac{528}{2^{32}}$

Step 5: Simplify the result. We simplify the fraction by factoring the numerator: $528 = 16 \times 33 = 2^4 \times 33$. Substitute this back into the probability expression: $\frac{2^4 \times 33}{2^{32}} = \frac{33}{2^{32 - 4}} = \frac{33}{2^{28}}$

This result corresponds to option (C).

3. Common Mistakes & Tips

• Invalid Parameters: Always check that the derived parameters $n$ and $p$ are valid ($n$ is a positive integer, $0 < p < 1$). Invalid parameters (like $q=2$) help rule out incorrect cases.
• Quadratic Equation Roots: When solving for mean and variance, remember that both are positive, which can sometimes help in selecting the correct pair of roots if one leads to invalid parameters.
• Simplification Errors: Be careful when simplifying fractions with powers of 2. $2^a / 2^b = 2^{a-b}$.

4. Summary

This problem required us to first determine the parameters of a binomial distribution ($n$ and $p$) using the given sum and product of its mean and variance. We formed a quadratic equation to find the mean and variance, then used the relationships $\mu = np$ and $\sigma^2 = npq$ to uniquely identify $n=32$ and $p=1/2$. Finally, we calculated the probability of one or two successes using the binomial probability mass function $P(X=k) = \binom{n}{k} p^k q^{n-k}$ and summed $P(X=1)$ and $P(X=2)$, simplifying the result to $\frac{33}{2^{28}}$.

5. Final Answer

The final answer is $\boxed{\frac{33}{2^{28}}}$, which corresponds to option (C).