1. Key Concepts and Formulas

• Binomial Distribution: This probability distribution is used when there is a fixed number of independent trials ($n$), each trial has only two possible outcomes (success or failure), and the probability of success ($p$) is constant for every trial. In this problem, answering each question is an independent trial.
• Probability Mass Function (PMF) of Binomial Distribution: The probability of getting exactly $k$ successes in $n$ trials is given by: $P(X=k) = {^n C_k} \cdot p^k \cdot (1-p)^{n-k}$ where:
  • $n$: Total number of trials (questions).
  • $k$: Number of successful outcomes (correct answers).
  • $p$: Probability of success in a single trial (guessing a question correctly).
  • $(1-p)$: Probability of failure in a single trial (guessing a question incorrectly).
  • ${^n C_k} = \frac{n!}{k!(n-k)!}$: The binomial coefficient, representing the number of ways to choose $k$ successes from $n$ trials.
• Addition Rule for Mutually Exclusive Events: When calculating the probability of event A or event B, and A and B cannot happen at the same time (mutually exclusive), we add their individual probabilities: $P(A \text{ or } B) = P(A) + P(B)$.

1. Step-by-Step Solution

Step 1: Identify the Parameters of the Binomial Distribution We begin by extracting the values for the total number of trials ($n$), the probability of success ($p$), and the probability of failure ($1-p$) directly from the problem statement.

• Total number of trials ($n$): The examination has $5$ questions. Therefore, $n=5$.
• Probability of success ($p$): Each question has three alternative answers, and exactly one is correct. If a student guesses randomly, the probability of selecting the correct answer for a single question is the ratio of correct options to total options. $p = P(\text{guessing correctly}) = \frac{\text{Number of correct options}}{\text{Total number of options}} = \frac{1}{3}$
• Probability of failure ($1-p$): The probability of guessing an incorrect answer for a single question is simply $1$ minus the probability of success. $1-p = P(\text{guessing incorrectly}) = 1 - \frac{1}{3} = \frac{2}{3}$

Step 2: Define the Event of Interest The problem asks for the probability that a student will get "4 or more correct answers" just by guessing. Since there are a total of 5 questions, "4 or more correct answers" means the student could get exactly 4 correct answers OR exactly 5 correct answers. Let $X$ be the random variable representing the number of correct answers. We need to find $P(X \ge 4)$. Since getting exactly 4 correct answers and exactly 5 correct answers are mutually exclusive events, we can use the addition rule: $P(X \ge 4) = P(X=4) + P(X=5)$

Step 3: Calculate the Probability of Exactly 4 Correct Answers ($P(X=4)$) We use the Binomial Distribution formula $P(X=k) = {^n C_k} \cdot p^k \cdot (1-p)^{n-k}$ with $n=5$, $k=4$, $p=1/3$, and $(1-p)=2/3$: $P(X=4) = {^5 C_4} \cdot \left(\frac{1}{3}\right)^4 \cdot \left(\frac{2}{3}\right)^{5-4}$ Let's break down each component:

• Binomial coefficient (${^5 C_4}$): This represents the number of ways to choose which 4 out of the 5 questions are answered correctly. ${^5 C_4} = \frac{5!}{4!(5-4)!} = \frac{5!}{4!1!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(1)} = 5$
• Probability of 4 successes $\left(\frac{1}{3}\right)^4$: This is the probability of getting 4 specific questions correct. $\left(\frac{1}{3}\right)^4 = \frac{1^4}{3^4} = \frac{1}{81}$
• Probability of 1 failure $\left(\frac{2}{3}\right)^1$: This is the probability of getting the remaining 1 question wrong. $\left(\frac{2}{3}\right)^1 = \frac{2}{3}$ Now, we multiply these components to find $P(X=4)$: $P(X=4) = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{5 \cdot 1 \cdot 2}{81 \cdot 3} = \frac{10}{243}$

Step 4: Calculate the Probability of Exactly 5 Correct Answers ($P(X=5)$) Next, we calculate $P(X=5)$ using the Binomial Distribution formula with $n=5$, $k=5$, $p=1/3$, and $(1-p)=2/3$: $P(X=5) = {^5 C_5} \cdot \left(\frac{1}{3}\right)^5 \cdot \left(\frac{2}{3}\right)^{5-5}$ Let's break down each component:

• Binomial coefficient (${^5 C_5}$): This is the number of ways to choose all 5 out of 5 questions to be answered correctly. There is only one way to achieve this. ${^5 C_5} = \frac{5!}{5!(5-5)!} = \frac{5!}{5!0!} = 1 \quad (\text{since } 0! = 1)$
• Probability of 5 successes $\left(\frac{1}{3}\right)^5$: This is the probability of getting all 5 questions correct. $\left(\frac{1}{3}\right)^5 = \frac{1^5}{3^5} = \frac{1}{243}$
• Probability of 0 failures $\left(\frac{2}{3}\right)^0$: Any non-zero number raised to the power of 0 is 1. $\left(\frac{2}{3}\right)^0 = 1$ Now, we multiply these components to find $P(X=5)$: $P(X=5) = 1 \cdot \frac{1}{243} \cdot 1 = \frac{1}{243}$

Step 5: Sum the Probabilities to Find the Final Answer Finally, to find the probability of getting 4 or more correct answers, we add the probabilities calculated in Step 3 and Step 4: $P(X \ge 4) = P(X=4) + P(X=5)$ $P(X \ge 4) = \frac{10}{243} + \frac{1}{243}$ Since both fractions already have a common denominator of 243 (which is $3^5$), we can directly add the numerators: $P(X \ge 4) = \frac{10 + 1}{243} = \frac{11}{243}$ This can be expressed as $\frac{11}{3^5}$.

1. Common Mistakes & Tips

• Misinterpreting "At Least/At Most": Carefully read phrases like "4 or more" ($X \ge 4$). This means you must sum the probabilities for all outcomes from 4 up to the maximum number of trials ($n$).
• Errors in Combinations: Ensure accurate calculation of the binomial coefficient ${^n C_k}$. Remember that $0! = 1$.
• Fractional Arithmetic: Be precise with powers of fractions and combining them, especially when finding a common denominator. In Binomial problems, the denominator is often $(\text{total outcomes per trial})^n$.

1. Summary

This problem is a standard application of the Binomial Distribution. We first identified the number of questions ($n=5$) and the probability of guessing a single question correctly ($p=1/3$). The event of interest, "4 or more correct answers," was broken down into two mutually exclusive cases: exactly 4 correct answers and exactly 5 correct answers. We calculated the probability for each case using the Binomial Probability Mass Function and then summed these probabilities to find the total probability of getting 4 or more correct answers. The calculated probability is $\frac{11}{243}$.

1. Final Answer

The final answer is $\boxed{\frac{11}{3^5}}$, which corresponds to option (C).