Solution

Let  $A$ and  $B$ denote the event that company is located in city  $A$ and  $B$, respectively.

$B$$A$

Let $F$ be the event that flexible schedule is available.

Let $n$ be the total number of observations and $n(\cdot)$ be the number of observations corresponding to an event.

a.

$$$$P(A) = \dfrac{n(A)}{n}= \dfrac{114}{220} \approx 0.52

b.

$$$$P(B \cap F)= \dfrac{n(B \cap F)}{n}= \dfrac{25}{220} \approx 0.11

c.

$$$$P(F^c)= \dfrac{n(F^c)}{n}= \dfrac{156}{220}\approx 0.71

d.

$$$$P(B| F)= \dfrac{P(B \cap F)}{P(F)}= \dfrac{n(B \cap F)}{n(F)}= \dfrac{25}{64}\approx 0.39

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