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$2$ take all three classes. What is the probability that a randomly selected student is taking exactly one language class?

Drawing a Venn Diagram does nothing and anyone who suggests that without further explanation should be punched in the face. This question is supposed to be answered using one of three formulas:

1) $P(E\cup F)=P(E)+P(F)-P(EF)$ and $P(E \cup F \cup G)= P(E)+P(F)+P(G)-P(EF)-P(EG)-P(FG)+P(EFG)$

2) $P(E)^c=1-P(E)$

3) $P(E)=\frac{\vert E \vert}{\vert S \vert}$

These were the only equations that we were shown in this chapter so as one user asked what is the probability that no one takes a class, well, it is precisely:

$P(E \cup F \cup G)^c=1-P(E \cup F \cup G)$

$=1-(P(E)+P(F)+P(G)-P(EF)-P(EG)-P(FG)+P(EFG)$

$=1-(.28+.26+.16-.12-.04-.06+.02)$

$=1-(.50)$

$=.50$

Now someone explain to me how this helps at all?

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    How many students are there in total? Or, said differently, are there any students that do not take any languages? – 2017-01-29
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    Draw a Venn Diagram – 2017-01-29
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    yes half do not take any languages – 2017-01-29

1 Answers 1

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Hint -

Draw venn diagram and start from intersection of all 3 events.

Edit -

page 1

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    From your suggestion I cam to realize the following: 1) The number of people taking Spanish only is equivalent to$ 28+(-4-2-12)$ 2) The number of people only taking German is equivalent to $16+(-4-2-6)$ and Finally 3) The number of people only taking only French is equal to $26+(-12-2-6)$ So now where do I go from here? – 2017-01-29
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    which leads to a total of 10 only taking Spanish, 6 only taking French, 4 only taking German. For a total of 20 students only taking exactly one class – 2017-01-29
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    Sorry I am sleeping that time. I uploaded one pic. If you have any doubt then ask. – 2017-01-30