# Probabilities in a multiple choice test

70 students take a multiple choice test with 100 questions. Each question has 4 possible answers— a, b, c, d — and there is only one correct answer per question. Student 1 gets 22 wrong answers and Student 2 gets 19 wrong answers. What is the probability that Students 1 and 2 have 13 wrong answers that are the same (eg. both picked “c” when c is one of the three possible wrong answers to a question) out of the 100 questions?

Does the fact that Student 1 had 13 out of 22 wrong answers that were the same as Student 2’s wrong answers prove that the two students were cheating off each other’s tests?

• This does not have an answer unless you assume that the students are using random guessing. Jun 21 '19 at 17:40
• Pretty high if there's "hot tip" going around that in this class you should pick (c) when in doubt because on the last few exams there were lots of correct (c) answers. Not unlikely that two students would believe it. Jun 21 '19 at 18:25

## 1 Answer

If all wrong answers are equally probable, it is indeed very unlikely that out of 19 wrong, 13 would match another's work. I calculate about 0.2%

$$p(i matches) = \left( \frac{1}{3} \right)^i \cdot \left( \frac{2}{3} \right) ^ {19-i} \cdot {{19}\choose{i}}$$ $$\sum_{i=13}^{19}{p(i)} \approx 0.0019$$

However!!!!!!! In every multiple choice test I've ever taken, there have always been better and worse distractor answers. And even just having one really obvious bad choice in each problem is enough to change the odds materially - I calculate better than 8% of at least 13 matches if there are only 2 good distractors. 8% might not sound high, but at 8% per student a class of 30 would have over 90% chance of at least one falling into that category!

• Very helpful-thank you! Using your assumption two distractors and an 8% chance per student, is it likely that 2 students would have 13 wrong answers the same? Can you calculate the probability? The class has 70 students.
– HMK
Jun 21 '19 at 21:41
• Thanks @HMK! Yes, as a matter of fact, I think I underestimated the 90%, treating every match as independent. Are you familiar with the Birthday Problem? The odds of two people sharing is less than 1/3 %, but all you need are 23 or 24 people for there to be a 50% chance of a match. I think this would be a similar case, so 8% per match with 70 students (assuming at least half had 19+ wrong out of 100) would nearly guarantee at least one such match. Jun 24 '19 at 11:50