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There are $25$ students in a class. $10$ of them are taking English course, $12$ are taking French course, $16$ German course, $6$ English and French, $7$ English and German, and $5$ of them are taking French and German course. We randomly choose one of the students. What is the probability that the chosen student is taking all three courses?

Note: This question may be similar to this post.

If we use inclusion-exclusion principle: $$|E\cup F\cup G|=|E|+|F|+|G|-|E\cap F|-|E\cap G|-|F\cap G|+|E\cap F\cap G|$$ where $$|E|=10,|F|=12,|G|=16,|E\cap F|=6,|E\cap G|=7,|F\cap G|=5$$

But, in order to solve the problem we must assume that every student is taking at least one course (since that is not specified).

Is this correct?

By this assumption we have that $|E\cup F\cup G|=25$, and the probability that the one chosen student is taking all three courses is $$P(E\cap F\cap G)=\frac{|E\cap F\cap G|}{|E\cup F\cup G|}=1/5$$

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    I did this roughly the same way. I have a more ham-fisted approach that uses Venn diagrams and doing calculations off to the side. I also got $1/5$. – 2017-01-11

1 Answers 1

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Is it possible for there to be a student that is taking no language?

Suppose that there is at least 1.

Then $|F\cup E\cup G| \le 24$

Which would then imply

$|F\cap E\cap G| \ge 6$

But $|F\cap G| = 5$ and that creates a contradiction.