In an examination of a certain class, at least $69\%$ of the students failed in P, at least $72\%$ failed in C, at least $80\%$ failed in M, and at least $85\%$ failed in B. How many at least must have failed in all the four subjects?

My attempt:

I'm familiar with the type of questions where the exact number of students is given for each set (like this). But in this question, the minimum number for each set is given instead!

Every attempt I make is leading me nowhere. I counted the number of students who have passed in each subject, but it still does not help.

I feel like I'm missing a specific train of thought that'll lead me to the correct answer. Thus, I need hints for realizing the idea of this problem. Thanks!


3 Answers 3


I like to look at the complements, since that's more intuitive: How many students can possibly have passed in at least one subject? We have that $31\%$ passed P, $28\%$ passed C, $20\%$ passed M and $15\%$ passed B. If we assume no overlap between these four groups for maximum at-least-one-subject-passed percentage, we get that at most $$ 31\% + 28\% + 20\% + 15\% = 94\% $$ passed on at least one exam (any overlap between the groups would've subtracted from this total). That means that at least $6\%$ failed on all exams.

  • $\begingroup$ This seems a lot simpler. $\endgroup$
    – Strawberry
    Jan 5, 2018 at 12:09
  • $\begingroup$ @Strawberry I think so too. All the subtracting done in all the other answers is really compensation for the fact that they're looking at the complements of the intuitively simple perspective. $\endgroup$
    – Arthur
    Jan 5, 2018 at 13:11

More generally, by Bonferroni inequality, $$P\left(\bigcup_{i=1}^NA_i^c\right)\leq \sum_{i=1}^NP(A_i^c),$$ that is $$P\left(\bigcap_{i=1}^NA_i\right)=1-P\left(\bigcup_{i=1}^NA_i^c\right)\ge1-\sum_{i=1}^NP(A_i^c)=\sum_{i=1}^NP(A_i)-(N-1).$$ In your case, $N=4$ and $$P\left(\bigcap_{i=1}^4A_i\right)\geq 0.69+0.72+0.80+0.85-(4-1)=0.06$$ where $A_1,A_2,A_3,A_4$ are the set of students who failed in P, C, M and B respectively.

  • $\begingroup$ Great! I didn't know the inequality, but I've understood what you've written. Thanks! $\endgroup$ Jan 4, 2018 at 11:49

The answer is 6%.

Firstly, we take any two subjects to compare. Let’s say we compare P and C. We have 69% failing P and 72% failing C. Hence, the minimum number of failures is 69%+72%-100%=41%.

Now, we compare M and B, with 80% and 85% failures respectively. Hence, the minimum number of failures is 80%+85%-100%=65%.

Now we compare those who failed P & C and those who failed M & B. The minimum number of failures of all 4 subjects is 65%+41%-100%=6%.

Sidenote: I believe these letters stand for physics, chemistry, biology and mathematics.

  • 1
    $\begingroup$ I'm not sure why this method of taking subjects in pairs works. You could also have added them all, then subtracted 300 to get the same answer. Then what exactly is the intuition behind pairing them and then calculating the final answer? Thanks! $\endgroup$ Jan 4, 2018 at 11:51
  • $\begingroup$ Yes, your method works too! I believe it is in fact the same. $\endgroup$
    – QuIcKmAtHs
    Jan 4, 2018 at 11:52
  • 2
    $\begingroup$ I'm just grouping in pairs as it feels easier to do, and also perhaps easier to understand. $\endgroup$
    – QuIcKmAtHs
    Jan 4, 2018 at 11:53
  • 1
    $\begingroup$ Using a venn diagram, you can see that if there are 69% failures at least, we maximise the number of non-failures, which is 100%-69% = 31%. Now 31% of these probably passed or failed(we don't know) but we want to minimise failures of both subjects. Now we can choose to put these 31% with those who did not fail the P, but those guys already have 1 non-failure, meaning that they will never be part of the failures. Hence, we rather give those who failed P this 31% so there are less complete failures. Hence it is 72%-31% = 41%. $\endgroup$
    – QuIcKmAtHs
    Jan 4, 2018 at 12:02
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    $\begingroup$ Yes, that is it! I used this approach as I thought you would be more familiar with the minimum of 2 subjects thing. I strongly recommend drawing a Venn diagram. $\endgroup$
    – QuIcKmAtHs
    Jan 4, 2018 at 12:17

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