Counting number of students who have failed in all four subjects

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.

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.

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.