Key Concepts and Formulas

• Binomial Probability Distribution: This distribution models the number of successes in a fixed number of independent Bernoulli trials, where each trial has only two possible outcomes (success or failure) and the probability of success remains constant.
• Binomial Probability Formula: 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 total number of trials, $k$ is the number of successes, $p$ is the probability of success in a single trial, and $q = 1-p$ is the probability of failure.
• Sum of Binomial Coefficients: For any positive integer $N$, the sum of all binomial coefficients is $2^N$: $\sum_{k=0}^N {^N C_k} = {^N C_0} + {^N C_1} + \dots + {^N C_N} = 2^N$

Step-by-Step Solution

Step 1: Identify Parameters and Formulate the Problem

First, let's define the parameters of our binomial distribution.

• Number of Trials ($N$): There are 8 true-false questions, so $N=8$.
• Probability of Success ($p$): The student guesses with equal probability for a true-false question, meaning there's a $1/2$ chance of guessing correctly. So, $p = {1 \over 2}$.
• Probability of Failure ($q$): The probability of guessing incorrectly is $q = 1 - p = 1 - {1 \over 2} = {1 \over 2}$.

Let $X$ be the random variable representing the number of correct answers. $X$ follows a Binomial Distribution, denoted as $X \sim B(8, {1 \over 2})$.

Using the Binomial Probability Formula, the probability of getting exactly $k$ correct answers is: $P(X=k) = {^8 C_k} \left({1 \over 2}\right)^k \left({1 \over 2}\right)^{8-k} = {^8 C_k} \left({1 \over 2}\right)^8 = {^8 C_k} \left({1 \over 256}\right)$

The problem asks for the smallest value of 'n' such that the probability of guessing at least 'n' correct answers is less than ${1 \over 2}$. "At least 'n' correct answers" translates to $X \ge n$. So, we need to find the smallest integer $n$ (where $0 \le n \le 8$) satisfying: $P(X \ge n) < {1 \over 2}$

Step 2: Express the Probability as a Sum of Binomial Coefficients

The probability $P(X \ge n)$ means the sum of probabilities of getting $n$, $n+1$, ..., up to $8$ correct answers. $P(X \ge n) = P(X=n) + P(X=n+1) + \dots + P(X=8)$ Substituting our simplified formula for $P(X=k)$: $P(X \ge n) = \sum_{k=n}^8 {^8 C_k} \left({1 \over 256}\right)$ Now, we incorporate this into our inequality from Step 1: $\sum_{k=n}^8 {^8 C_k} \left({1 \over 256}\right) < {1 \over 2}$ To simplify, we multiply both sides by $256$: $\sum_{k=n}^8 {^8 C_k} < {1 \over 2} \times 256$ $\sum_{k=n}^8 {^8 C_k} < 128$ This inequality states that the sum of binomial coefficients from ${^8 C_n}$ to ${^8 C_8}$ must be less than 128.

Step 3: Utilize the Property of Complementary Sums of Binomial Coefficients

We know that the sum of all binomial coefficients for $N=8$ is $2^8 = 256$: $\sum_{k=0}^8 {^8 C_k} = {^8 C_0} + {^8 C_1} + \dots + {^8 C_8} = 256$ We can express the sum $\sum_{k=n}^8 {^8 C_k}$ using the total sum: $\sum_{k=n}^8 {^8 C_k} = \left(\sum_{k=0}^8 {^8 C_k}\right) - \left(\sum_{k=0}^{n-1} {^8 C_k}\right)$ Substituting this into our inequality $\sum_{k=n}^8 {^8 C_k} < 128$: $256 - \left({^8 C_0} + {^8 C_1} + \dots + {^8 C_{n-1}}\right) < 128$ Now, we rearrange the inequality to isolate the sum we need to evaluate: $256 - 128 < {^8 C_0} + {^8 C_1} + \dots + {^8 C_{n-1}}$ $128 < \sum_{k=0}^{n-1} {^8 C_k}$ Our goal is to find the smallest integer $n$ for which the sum of binomial coefficients from ${^8 C_0}$ up to ${^8 C_{n-1}}$ is greater than 128.

Step 4: Evaluate Binomial Coefficients and Determine 'n'

Let's calculate the required binomial coefficients for $N=8$:

• ${^8 C_0} = 1$
• ${^8 C_1} = 8$
• ${^8 C_2} = \frac{8 \times 7}{2 \times 1} = 28$
• ${^8 C_3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
• ${^8 C_4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$

Now, we accumulate the sum $S_{n-1} = \sum_{k=0}^{n-1} {^8 C_k}$ and check the condition $S_{n-1} > 128$:

• For $n=1$ ($n-1=0$): $S_0 = {^8 C_0} = 1$. Is $1 > 128$? No.
• For $n=2$ ($n-1=1$): $S_1 = {^8 C_0} + {^8 C_1} = 1 + 8 = 9$. Is $9 > 128$? No.
• For $n=3$ ($n-1=2$): $S_2 = S_1 + {^8 C_2} = 9 + 28 = 37$. Is $37 > 128$? No.
• For $n=4$ ($n-1=3$): $S_3 = S_2 + {^8 C_3} = 37 + 56 = 93$. Is $93 > 128$? No.
• For $n=5$ ($n-1=4$): $S_4 = S_3 + {^8 C_4} = 93 + 70 = 163$. Is $163 > 128$? Yes!

The condition $128 < \sum_{k=0}^{n-1} {^8 C_k}$ is first satisfied when $n-1 = 4$, which means $n=5$. Therefore, the smallest value of $n$ is 5.

Common Mistakes & Tips

• Misinterpreting "at least n": A common mistake is to confuse $P(X \ge n)$ with $P(X=n)$ (exactly n correct answers) or $P(X \le n)$ (at most n correct answers). Always remember that "at least n" implies summing probabilities for $k=n, n+1, \dots, N$.
• Calculation Errors: Carefully compute binomial coefficients. For smaller $N$, direct calculation is fine. Remember the symmetry property: ${^N C_k} = {^N C_{N-k}}$.
• Symmetry of Binomial Distribution (for $p=1/2$): When $p=1/2$, the binomial distribution is symmetric around its mean $N/2$. For $N=8$, the mean is 4. This means $P(X \ge 4)$ would naturally be $\ge 1/2$.
  • $\sum_{k=0}^8 {^8 C_k} = 256$.
  • $\sum_{k=0}^3 {^8 C_k} = 1+8+28+56 = 93$.
  • By symmetry, $\sum_{k=5}^8 {^8 C_k} = \sum_{k=0}^3 {^8 C_k} = 93$.
  • $P(X \ge 5) = \frac{93}{256}$. Is $\frac{93}{256} < \frac{1}{2}$? Yes, because $93 < 128$. So $n=5$ works.
  • $P(X \ge 4) = P(X=4) + P(X \ge 5) = \frac{{^8 C_4}}{256} + \frac{93}{256} = \frac{70+93}{256} = \frac{163}{256}$. Is $\frac{163}{256} < \frac{1}{2}$? No, because $163 > 128$. So $n=4$ does not work. This confirms $n=5$ is the smallest value.
• Inequality Direction: Pay close attention to the direction of the inequality sign when rearranging terms, especially when multiplying or dividing by negative numbers (though not applicable in this problem).

Summary

This problem is a standard application of the Binomial Probability Distribution. We first identified the distribution parameters ($N=8$, $p=1/2$) and formulated the problem as an inequality involving the probability of "at least $n$" correct answers. By expressing this probability as a sum of binomial coefficients and leveraging the property of the total sum of coefficients, we transformed the inequality into a condition on a partial sum of coefficients. Systematically evaluating these sums revealed that $n=5$ is the smallest integer satisfying the given condition.

The final answer is $\boxed{5}$, which corresponds to option (A).