There are $25$ students in one class. $10$ of them are studying English, $12$ French, $16$ German. $6$ are studying both English and French, $7$ are studying English and German and $5$ are studying French and German.
We choose one student randomly. What is the probability that they are studying all $3$ languages?
I tried it like this:
A - students that are studying English
B - students that are studying French
C - students that are studying German
$P(A \cap B \cap C) = 1-P(A^c\cup B^c\cup C^c)= 1-[P(A^c\cup B^c)+P(C^{c})-P((A^c\cup B^c)\cap C^c)] = 1-[P((A\cap B)^c)+P(C^{c})-(1-P((A\cap B)\cup C))]$
But now I don't know what to do with $P((A\cap B)\cup C))$. If I do the union thing again I will just be back at the beginning.
Draw a Venn diagram. For the triple intersection put in your unknown a. Now as EUF=6, for EUFUG' put in 6-a. In similar fashion, FUGUE'=5-a; & EUGUF'=7-a.
But now, EUF'UG' becomes 10-[6+(7-a)]; FUE'UG'=12-[6+(5-a)]; & GUE'UF'=16-[5+(7-a)].
But now we have a problem: is it assumed that everyone in the class is taking at least one language?! For o/w we have 2 unknowns where a+b=5.
If all are taking at least one, then a=5, & the probability will be 5/25=1/5.