Of the $40$ employees at a certain company, $20$ are available to meet on Monday $(M)$, $17$ are available to meet on wednesday $(W)$, and $8$ are available neither on Monday nor wednesday $(\overline{M\cup W})$. What is the probability that a randomly selected employee is available to meet on both Monday and Wednesday $(M \cap W)$?
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$\begingroup$Let $M$ be the set of people available on monday and $W$ be the set of people available on wednesday.$$P(M)=\dfrac{20}{40}$$$$P(W)=\dfrac{17}{40}$$Now $P(\overline{M\cup W})=\dfrac{8}{40}$$$\Rightarrow P(M\cup W)=\dfrac{32}{40}$$$$P(M\cap W)=P(M)+P(W)-P(M\cup W)$$$$P(M\cap W)=\dfrac{20}{40}+\dfrac{17}{40}-\dfrac{32}{40}$$$$P(M\cap W)=\dfrac{1}{8}$$Therefore the probability that a randomly selected employee is available on both monday and wednesday is $\dfrac{1}{8}$
$\endgroup$ $\begingroup$Let $E$ denote the set of employees.
Let $M\subseteq E$ denote the set of employees available for monday.
Let $W\subseteq E$ denote the set of employees available for wednesday.
Now use the equality:$$|W|+|M|=|W\cup M|+|W\cap M|$$
To get some insight make a Venn diagram.
$\endgroup$ $\begingroup$Hint:
$$40 =\underbrace{|M| +|W| -|M\cap W|}_{|M\cup W|} + |(M\cup W)^c|$$
This is nothing more than the statement that the Venn diagram includes all employees.
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