1. Key Concepts and Formulas

• Binomial Distribution Probability Mass Function (PMF): For a random variable $X$ following a binomial distribution $B(n, p)$, the probability of getting exactly $k$ successes in $n$ trials is given by: $P(X=k) = {}^n C_k p^k q^{n-k}$ 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 ${}^n C_k = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.
• Mean of Binomial Distribution: The expected value (mean) of a binomial distribution is given by: $\mu = np$
• Variance of Binomial Distribution: The variance of a binomial distribution is given by: $\sigma^2 = npq$

2. Step-by-Step Solution

Step 1: Extract Given Information and Formulate Equations We are given the mean and variance of a binomial distribution. Our first goal is to use these to determine the parameters $n$ (number of trials) and $p$ (probability of success in a single trial), and consequently $q$ (probability of failure).

• Given Mean ($\mu$): $\mu = 4$
• Given Variance ($\sigma^2$): $\sigma^2 = 2$

Using the standard formulas for mean and variance of a binomial distribution, we can write:

1. $np = 4 \quad (\text{Equation 1})$
2. $npq = 2 \quad (\text{Equation 2})$

Step 2: Determine the Probability of Failure ($q$) To find $q$, we can divide Equation 2 by Equation 1. This is a common strategy as the $np$ term will cancel out, simplifying the expression directly to $q$. $\frac{npq}{np} = \frac{2}{4}$ $q = \frac{1}{2}$ This means the probability of failure in a single trial is $1/2$.

Step 3: Determine the Probability of Success ($p$) The sum of the probability of success ($p$) and the probability of failure ($q$) must always be $1$. $p + q = 1$ Substitute the value of $q$ we just found: $p + \frac{1}{2} = 1$ $p = 1 - \frac{1}{2}$ $p = \frac{1}{2}$ So, the probability of success in a single trial is also $1/2$.

Step 4: Determine the Number of Trials ($n$) Now that we have the value of $p$, we can substitute it back into Equation 1 ($np=4$) to solve for $n$. $n \left(\frac{1}{2}\right) = 4$ To isolate $n$, multiply both sides of the equation by $2$: $n = 4 \times 2$ $n = 8$ This means there are $8$ trials in this binomial distribution.

Step 5: Identify the Parameters of the Binomial Distribution We have successfully determined all the necessary parameters for our binomial distribution:

• Number of trials ($n$) = $8$
• Probability of success ($p$) = $\frac{1}{2}$
• Probability of failure ($q$) = $\frac{1}{2}$

Step 6: Calculate the Probability of 2 Successes The question asks for the probability of $2$ successes. Using the PMF formula $P(X=k) = {}^n C_k p^k q^{n-k}$ with $n=8$, $p=\frac{1}{2}$, $q=\frac{1}{2}$, and $k=2$: $P(X=2) = {}^8 C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{8-2}$ $P(X=2) = {}^8 C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^6$

Step 7: Compute the Binomial Coefficient ${}^8 C_2$ ${}^8 C_2 = \frac{8!}{2!(8-2)!} = \frac{8!}{2!6!}$ Expand the factorials: ${}^8 C_2 = \frac{8 \times 7 \times 6!}{2 \times 1 \times 6!}$ Cancel out $6!$: ${}^8 C_2 = \frac{8 \times 7}{2} = \frac{56}{2} = 28$

Step 8: Compute the Powers of $p$ and $q$ Since $p=1/2$ and $q=1/2$, we can combine the powers: $\left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^6 = \left(\frac{1}{2}\right)^{2+6} = \left(\frac{1}{2}\right)^8$ Now, calculate the value of $\left(\frac{1}{2}\right)^8$: $\left(\frac{1}{2}\right)^8 = \frac{1^8}{2^8} = \frac{1}{256}$

Step 9: Combine the Results for the Final Probability Substitute the calculated values from Step 7 and Step 8 back into the expression for $P(X=2)$: $P(X=2) = 28 \times \frac{1}{256}$ $P(X=2) = \frac{28}{256}$

3. Common Mistakes & Tips

• Formula Recall: Ensure you have the formulas for mean, variance, and the PMF of a binomial distribution memorized. These are fundamental.
• Systematic Parameter Finding: Always prioritize finding $n, p,$ and $q$ first. Rushing to calculate probabilities without correctly identifying these parameters is a common error.
• Algebraic Accuracy: Pay close attention to algebraic manipulations, especially when solving simultaneous equations (like dividing $npq$ by $np$) and handling fractions and exponents.
• Binomial Coefficient Calculation: Practice calculating ${}^n C_k$ efficiently. Remember the shortcut ${}^n C_k = \frac{n(n-1)...(n-k+1)}{k!}$. For example, ${}^8 C_2 = \frac{8 \times 7}{2 \times 1}$.

4. Summary

This problem required us to apply the core concepts of a binomial distribution. We started by using the given mean and variance to systematically solve for the distribution's parameters: the number of trials ($n=8$), the probability of success ($p=1/2$), and the probability of failure ($q=1/2$). Once these parameters were established, we used the Probability Mass Function (PMF) to calculate the probability of exactly 2 successes, which involved computing a binomial coefficient and powers of fractions. The final calculated probability is $\frac{28}{256}$.

5. Final Answer

The probability of $2$ successes is $\boxed{\frac{28}{256}}$, which corresponds to option (A).