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?

  • 1
    $\begingroup$ 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 $\endgroup$
    – user208649
    Aug 19, 2020 at 23:19
  • $\begingroup$ @TokenToucan I think that may be right. Is there a way to quantify this idea? $\endgroup$
    – Newman
    Aug 19, 2020 at 23:22

3 Answers 3


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 $0.5<p<2/3$ 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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – user208649
    Aug 19, 2020 at 22:30
  • $\begingroup$ A good point, if we're using coin flips. Then the correct answer uses binomial coefficients. Will edit. $\endgroup$ Aug 19, 2020 at 22:37
  • $\begingroup$ Edited, thanks for the correction. $\endgroup$ 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.


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