Question

Is it possible for the teacher to detect cheating without monitoring? Choose the statement that best describes your opinion:

• A. It is not at all possible; teacher will have to introduce technology if there is no human support.
• B. It is always possible, but teacher has to calculate the exact answer.
• C. It is possible when many students sitting next to each other have the same incorrect answers for multiple questions. However, there can be a small error in judgment.
• D. It is possible when many students sitting next to each other have the same correct answers for multiple questions. However, there can be a small error in judgment.
• E. It is possible only for poor students but not for good and brilliant students. However, there can be a small error in judgment.
• A. 256/390625
• B. 256/3125
• C. 4/3125
• D. 1/3125
• J. Cannot be calculated

Answer 1

The Correct Option is C

Step 1: Analyze the core of the question.
The question asks for the best method to detect cheating by analyzing answer sheets after a quiz, without direct monitoring during the quiz. This relies on finding statistical anomalies.
Step 2: Evaluate the options based on statistical probability.
(A) Not at all possible: This is an absolute statement and is incorrect. Statistical analysis of answer patterns is a valid method for detecting cheating.
(B) Always possible: This is also an absolute. Detecting cheating this way is about probability, not certainty. The teacher knowing the right answer is for grading, not directly for detecting cheating.
(C) Same incorrect answers: This is a very strong indicator of cheating. In a multiple-choice question with six options, there is one correct way to answer but five incorrect ways. It is statistically highly improbable for two students, sitting together, to independently choose the exact same wrong answer for multiple questions. This shared pattern of errors is a classic red flag. The caveat "there can be a small error in judgment" is accurate because it's a statistical inference, not absolute proof.
(D) Same correct answers: This is a much weaker indicator. It is plausible, especially for "good" or "brilliant" students, to arrive at the same correct answers through their own knowledge. While suspicious, it's not as statistically unlikely as sharing multiple identical errors.
(E) Only for poor students: This is a flawed assumption. Any student, regardless of ability, can cheat. The detection method based on answer patterns is not dependent on the students' academic level.
Step 3: Conclude the most logical method.
The most reliable and scientifically sound method for detecting cheating after the fact is to identify pairs or groups of students with an improbably high number of identical incorrect answers.
Therefore: The best statement is that it is possible when students share the same incorrect answers for multiple questions. \[ \boxed{\text{(C)}} \]

Answer 2

Step 1: Identify the given probabilities.
Number of answer choices per question = 6 (1 correct, 5 incorrect).
For a "good student," the probability of answering a question correctly is P(Correct) = 0.2.
Therefore, the probability of a "good student" answering a question incorrectly is P(Incorrect) = 1 - 0.2 = 0.8.
Step 2: Identify the missing information.
The question asks for the probability of the three students selecting the same incorrect choice. While we know the overall probability of a student being incorrect is 0.8, we do not know how this probability is distributed among the 5 incorrect answer choices.
Step 3: Analyze the problem's ambiguity.
In a well-designed multiple-choice question, the incorrect options (distractors) are not all equally plausible. Some are designed to be more tempting than others. A "good student" who answers incorrectly is more likely to choose a plausible distractor than a completely random wrong answer. The problem provides no information about the probability of selecting any *specific* incorrect answer. We cannot assume that the 0.8 probability of being wrong is split evenly among the 5 wrong choices (i.e., 0.16 for each).
Step 4: Conclude the feasibility of calculation.
To calculate the probability of all three students choosing the same incorrect answer, we would need to know P(student chooses incorrect option X | student answers incorrectly). Since this information is not provided and cannot be reasonably assumed, a precise calculation is impossible.
Therefore: The probability cannot be calculated from the given information. \[ \boxed{\text{Cannot be calculated}} \]