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 and Variance: For a binomial distribution $B(n,p)$, the mean (expected value) is $E[X] = np$, and the variance is $Var[X] = npq$.
• Probability Mass Function (PMF): The probability of getting exactly $r$ successes in $n$ trials is given by $P(X=r) = {^nC_r} p^r q^{n-r}$, for $r = 0, 1, \ldots, n$.

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Step-by-Step Solution

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

The first crucial step is to identify the parameters $n$ (number of trials) and $p$ (probability of success) that define this specific binomial distribution. We are given the mean and variance, which allows us to find these parameters.

We are given:

• Mean ($E[X]$) = 8
• Variance ($Var[X]$) = 4

Using the formulas for the mean and variance of a binomial distribution: $np = 8 \quad \text{(Equation 1)}$ $npq = 4 \quad \text{(Equation 2)}$

To find $q$, we can divide Equation 2 by Equation 1. This eliminates $n$ and $p$, directly giving us $q$: $\frac{npq}{np} = \frac{4}{8}$ $q = \frac{1}{2}$

Now that we have $q$, we can find $p$ using the fundamental relationship $p+q=1$: $p = 1 - q$ $p = 1 - \frac{1}{2}$ $p = \frac{1}{2}$

Finally, we substitute the value of $p$ back into Equation 1 to find $n$: $n \left(\frac{1}{2}\right) = 8$ $n = 8 \times 2$ $n = 16$

Thus, the random variable $X$ follows a binomial distribution $B(16, \frac{1}{2})$. This means there are 16 trials, and the probability of success in each trial is $1/2$.

Step 2: Formulate the Probability Mass Function (PMF) for this specific distribution

With the parameters $n=16$, $p=\frac{1}{2}$, and $q=\frac{1}{2}$, we can write the specific PMF for $X$. This formula will be used to calculate the probability of any number of successes $r$.

The general PMF is $P(X=r) = {^nC_r} p^r q^{n-r}$. Substituting our values: $P(X=r) = {^{16}C_r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{16-r}$ Since the base is the same, we can combine the exponential terms: $P(X=r) = {^{16}C_r} \left(\frac{1}{2}\right)^{r + (16-r)}$ $P(X=r) = {^{16}C_r} \left(\frac{1}{2}\right)^{16}$ This can be written more compactly as: $P(X=r) = \frac{^{16}C_r}{2^{16}}$

Step 3: Calculate the Required Probability $P(X \le 2)$

The problem asks for $P(X \le 2)$, which means the probability that the number of successes is less than or equal to 2. For a binomial distribution, this involves summing the probabilities for $X=0$, $X=1$, and $X=2$: $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$

Using our derived PMF $P(X=r) = \frac{^{16}C_r}{2^{16}}$:

• For $X=0$ successes ($r=0$): $P(X=0) = \frac{^{16}C_0}{2^{16}} = \frac{1}{2^{16}} \quad \left(\text{since } {^nC_0} = 1\right)$

• For $X=1$ success ($r=1$): $P(X=1) = \frac{^{16}C_1}{2^{16}} = \frac{16}{2^{16}} \quad \left(\text{since } {^nC_1} = n\right)$

• For $X=2$ successes ($r=2$): $P(X=2) = \frac{^{16}C_2}{2^{16}}$ First, calculate the binomial coefficient: ${^{16}C_2} = \frac{16 \times (16-1)}{2 \times 1} = \frac{16 \times 15}{2} = 8 \times 15 = 120$ So, $P(X=2) = \frac{120}{2^{16}}$

Now, sum these probabilities: $P(X \le 2) = \frac{1}{2^{16}} + \frac{16}{2^{16}} + \frac{120}{2^{16}}$ $P(X \le 2) = \frac{1 + 16 + 120}{2^{16}}$ $P(X \le 2) = \frac{137}{2^{16}}$

Step 4: Compare with the Given Form and Find $k$

The problem states that $P(X \le 2) = \frac{k}{2^{16}}$. We calculated $P(X \le 2) = \frac{137}{2^{16}}$.

By comparing these two expressions, we can directly find the value of $k$: $\frac{k}{2^{16}} = \frac{137}{2^{16}}$ $k = 137$

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Common Mistakes & Tips

• Incorrect Parameter Derivation: A common error is miscalculating $n$ or $p$ from the mean and variance. Always double-check by dividing $npq$ by $np$ to find $q$, then use $p=1-q$, and finally $n=E[X]/p$.
• Misinterpreting Probability Notation: Be careful to distinguish between $P(X=r)$ (exactly $r$ successes) and $P(X \le r)$ (at most $r$ successes). The latter requires summing individual probabilities.
• Binomial Coefficient Errors: Ensure correct calculation of ${^nC_r}$, especially for common cases like ${^nC_0}=1$, ${^nC_1}=n$, and ${^nC_2}=\frac{n(n-1)}{2}$.

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Summary

This problem required us to first determine the underlying binomial distribution parameters ($n$ and $p$) using the given mean and variance. We found $n=16$ and $p=1/2$. With these parameters, we formulated the specific probability mass function (PMF). The final step involved calculating the cumulative probability $P(X \le 2)$ by summing the probabilities of $X=0$, $X=1$, and $X=2$ using the PMF. Comparing this result with the given form $\frac{k}{2^{16}}$ allowed us to find the value of $k$.

The final answer is $\boxed{\text{137}}$, which corresponds to option (C).