1. Key Concepts and Formulas

• Binomial Distribution $B(n, p)$: A random variable $X$ follows a binomial distribution if it represents the number of successes 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 ($\mu$): $\mu = np$
  • Variance ($\sigma^2$): $\sigma^2 = npq$
• Probability Mass Function (PMF): The probability of exactly $x$ successes in $n$ trials is given by: $P(X=x) = {^nC_x} p^x q^{n-x}$ where $x \in \{0, 1, \dots, n\}$.
• Properties of Combinations: ${^nC_x} = \frac{n!}{x!(n-x)!}$. An important property is ${^nC_x} = {^nC_{n-x}}$.

2. Step-by-Step Solution

Step 1: Formulate equations using the given information about mean and variance. We are given that the sum of the mean and variance is 24, and their product is 128. Let $\mu$ be the mean and $\sigma^2$ be the variance of $X$.

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

Now, substitute the formulas for mean ($\mu = np$) and variance ($\sigma^2 = npq$) into these equations:

• Equation (1): $np + npq = 24$
• Equation (2): $(np)(npq) = 128$

Step 2: Solve the system of equations to find the values of $np$ and $q$. To simplify, let $M = np$ (which represents the mean). The equations become:

• $M + Mq = 24$
• $M \cdot (Mq) = 128$

From the first equation, we can express $Mq$ in terms of $M$: $Mq = 24 - M$ Substitute this expression for $Mq$ into the second equation: $M(24 - M) = 128$ Rearrange this into a standard quadratic equation: $24M - M^2 = 128$ $M^2 - 24M + 128 = 0$ We can solve this quadratic equation by factoring. We look for two numbers that multiply to 128 and add up to -24. These numbers are -16 and -8. $(M - 16)(M - 8) = 0$ This gives two possible values for $M$ (the mean $np$): $M = 16 \quad \text{or} \quad M = 8$

Step 3: Determine the valid parameters ($n, p, q$) for the binomial distribution. We must check both possible values for $M = np$.

• Case 1: If $M = np = 16$ Substitute $M=16$ into $Mq = 24 - M$: $16q = 24 - 16$ $16q = 8$ $q = \frac{8}{16} = \frac{1}{2}$ Since $p + q = 1$, we find $p$: $p = 1 - q = 1 - \frac{1}{2} = \frac{1}{2}$ Now, use $np = 16$ to find $n$: $n \left(\frac{1}{2}\right) = 16$ $n = 32$ Validity Check: $n=32$ is a positive integer, and $p=1/2$, $q=1/2$ are probabilities between 0 and 1. This set of parameters is valid.

• Case 2: If $M = np = 8$ Substitute $M=8$ into $Mq = 24 - M$: $8q = 24 - 8$ $8q = 16$ $q = \frac{16}{8} = 2$ Validity Check: The probability $q$ must be between 0 and 1. Since $q=2$ is greater than 1, this case is invalid.

Therefore, the only valid parameters for the binomial distribution are $n=32$, $p=1/2$, and $q=1/2$.

Step 4: Calculate the probability $P(X > n-3)$. Using $n=32$, we need to calculate $P(X > 32-3)$, which simplifies to $P(X > 29)$. Since $X$ is a discrete random variable taking integer values from $0$ to $n$, $P(X > 29)$ means $P(X=30) + P(X=31) + P(X=32)$.

The Probability Mass Function (PMF) is $P(X=x) = {^nC_x} p^x q^{n-x}$. With $n=32$, $p=1/2$, $q=1/2$, the term $p^x q^{n-x}$ becomes $(1/2)^x (1/2)^{32-x} = (1/2)^{32}$. So, $P(X=x) = {^{32}C_x} \left(\frac{1}{2}\right)^{32}$.

Let's calculate each term:

• For $X=30$: $P(X=30) = {^{32}C_{30}} \left(\frac{1}{2}\right)^{32}$ Using ${^nC_x} = {^nC_{n-x}}$, we have ${^{32}C_{30}} = {^{32}C_{32-30}} = {^{32}C_2}$. ${^{32}C_2} = \frac{32 \times 31}{2 \times 1} = 16 \times 31 = 496$ So, $P(X=30) = 496 \left(\frac{1}{2}\right)^{32}$.

• For $X=31$: $P(X=31) = {^{32}C_{31}} \left(\frac{1}{2}\right)^{32}$ Using ${^nC_x} = {^nC_{n-x}}$, we have ${^{32}C_{31}} = {^{32}C_{32-31}} = {^{32}C_1}$. ${^{32}C_1} = 32$ So, $P(X=31) = 32 \left(\frac{1}{2}\right)^{32}$.

• For $X=32$: $P(X=32) = {^{32}C_{32}} \left(\frac{1}{2}\right)^{32}$ ${^{32}C_{32}} = 1$ So, $P(X=32) = 1 \left(\frac{1}{2}\right)^{32}$.

Now, sum these probabilities: $P(X > 29) = 496 \left(\frac{1}{2}\right)^{32} + 32 \left(\frac{1}{2}\right)^{32} + 1 \left(\frac{1}{2}\right)^{32}$ Factor out the common term $\left(\frac{1}{2}\right)^{32}$: $P(X > 29) = (496 + 32 + 1) \left(\frac{1}{2}\right)^{32}$ $P(X > 29) = 529 \left(\frac{1}{2}\right)^{32}$

Step 5: Determine the value of $k$. The problem states that $P(X > n-3) = \frac{k}{2^n}$. Substituting $n=32$, we have $P(X > 29) = \frac{k}{2^{32}}$. Comparing this with our calculated probability: $\frac{k}{2^{32}} = 529 \left(\frac{1}{2}\right)^{32} = \frac{529}{2^{32}}$ Therefore, $k = 529$.

3. Common Mistakes & Tips

• Parameter Validity: Always verify that the derived parameters $n, p, q$ are valid for a binomial distribution ($n$ is a positive integer, $0 \le p \le 1$, $0 \le q \le 1$). This step is crucial for eliminating extraneous solutions from quadratic equations.
• Inequality Interpretation: For discrete random variables, $P(X > a)$ means $P(X=a+1) + P(X=a+2) + \dots$. Ensure you correctly identify the range of values for summation.
• Combinations and PMF Simplification: When $p=q=1/2$, the term $p^x q^{n-x}$ simplifies to $(1/2)^n$, which can significantly streamline calculations. Also, use the property ${^nC_x} = {^nC_{n-x}}$ to simplify calculations for large $x$.

4. Summary

This problem tests the fundamental understanding of binomial distribution properties. It involved setting up and solving a system of equations based on the given sum and product of mean and variance to determine the distribution parameters ($n=32, p=1/2, q=1/2$). A critical step was validating the parameters to discard invalid solutions. Finally, the probability $P(X > n-3)$ was calculated by summing the probabilities of specific outcomes using the PMF, leading to the value of $k$.

The final answer is $\boxed{\text{529}}$, which corresponds to option (B).