# Probability of passing an exam given two different grading schemes

I had a discussion with a peer about the probability of passing an exam given two different grading schemes and I'm not sure I believe what my peer stated.

They stated that given an exam with 10 questions where a passing grade is 5 correct questions and an exam with 6 questions where a passing grade is 3 questions, it would be better to take the exam with 10 questions as there is an increased probability of passing. I didn't buy the argument as it seems that the exams are equivalent, i.e., you need a 50% to pass either. However my peer was adamant about the their point. Can anyone clarify this?

• My guess is your friends intuition is something like: on a 10 question exam, I stand a better chance of being familiar with at least half of the topics and hence of getting those questions right; on a 6 question exam there is less of a topic spread which risks not having enough familiari
– user208649
Aug 19, 2020 at 23:19
• @TokenToucan I think that may be right. Is there a way to quantify this idea? Aug 19, 2020 at 23:22

The rightness or wrongness of your peer's statement depends on the probability of success of answering each question correctly.

If you assume that the test is a set of $$2N$$ true/false questions, with N correct answers required to pass, where your probability of answering any question is $$p$$, then the probability $$P$$ of passing the test is such that:

for $$p<0.5$$, $$P$$ falls monotonically with increasing N and in the limit of $$N {\rightarrow} {\infty}$$, $$P {\rightarrow} 0$$, so it will always be preferential to choose the test with the least number of questions.

for $$p=0.5$$ the probability of passing still falls with increasing N (but now asymptotes to 0.5), $$N {\rightarrow} {\infty}$$, $$P {\rightarrow} 0.5$$, so still choose the test with the least number of questions.

for $$0.5 the probability of passing initially falls with increasing N, but then increases with larger N and in the limit $$N {\rightarrow} {\infty}$$, $$P {\rightarrow} 1.0$$, so your choice would depend on the maximum number of questions. For example, if $$p=0.51$$ then sitting a test with $$N\simeq570$$ questions is marginally better than sitting a test with $$N=2$$ questions.

for $$p>2/3$$ the probability of passing increases monotonically with increasing N, and in the limit $$N {\rightarrow} {\infty}$$, $$P {\rightarrow} 1.0$$, so you should always choose the test with the most questions.

In your example, choosing either a 6 question or a 10 question test, your probability of success will be approximately equal if $$p\simeq0.564$$ (in that case $$P\simeq0.7674$$), it would be better to do the 6 question test if $$p<0.564$$, but you should choose the 10 question test if $$p>0.564$$.

Your friend is incorrect, assuming you're flipping a coin to determine whether you're right or wrong. You've both forgotten that a score of 0 is possible, which means the chance of passing isn't 50% for either.

On the test with 10 questions, 6 of 11 possible scores pass. On the 6 question test, 4/7 scores pass. If you're tossing coins, then the probabilities of passing are

$$\frac{1}{2^6}\sum_0^3 {6 \choose k}$$ or $$\frac{1}{2^{10}}\sum_0^6 {10 \choose k}$$

For the six and 10 question tests respectively. That's 65% for the 6 question test, and 62% for the 10 question test.

It's better to take the 6-question exam if you truly think each question is a coin toss as to whether you're right.

• Not all scores are equally likely. Your method would say a probability of $2/3$ for passing a two question exam when it is actually $3/4$. The difference is one can get just 1 question right in two different ways.
– user208649
Aug 19, 2020 at 22:30
• A good point, if we're using coin flips. Then the correct answer uses binomial coefficients. Will edit. Aug 19, 2020 at 22:37
• Edited, thanks for the correction. Aug 19, 2020 at 22:46

Answer: it depends, but for a true/false test, your peer is incorrect.

Getting 5 right and 5 wrong is how to order RRRRRWWWWW where R is right and W is wrong. There are $$\binom{10}{5} = 252$$ ways. Similarly for 6 right, 7 right, etc.

This is $$\sum_{i=0}^{5} \binom{10}{i}$$, which equals $$638$$. Dividing by $$2^{10}$$ gets us approx. $$0.623$$.

Now, getting 3 right and 3 wrong is ordering RRRWWW. We want to find $$\sum_{i=0}^{3} \binom{6}{i}$$, which comes out to $$42$$. Dividing by $$2^6$$ gets us approx. $$0.652$$.

$$0.623 < 0.652$$, so your peer is incorrect.