Counting number of students who have failed in all four subjects 
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!
 A: 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.
A: 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.
A: 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.
