Suppose p(e) = 57⁄100 , p(fc ) = 7⁄20 , and p(f ∩ ec ) = 31⁄100. find p(e ∪ f).

By the inclusion/exclusion principle,

[tex]\mathbb P(E\cup F)=\mathbb P(E)+\mathbb P(F)-\mathbb P(E\cap F)[/tex]

Recall the law of total probability:

[tex]\mathbb P(F)=\mathbb P(E\cap F)+\mathbb P(E^C\cap F)[/tex]

We're given

[tex]P(F^C)=\dfrac7{20}\implies\mathbb P(F)=1-\mathbb P(F^C)=\dfrac{13}{20}[/tex]

and so

[tex]\dfrac{13}{20}=\mathbb P(E\cap F)+\dfrac{31}{100}\implies\mathbb P(E\cap F)=\dfrac{17}{50}[/tex]

Now,

[tex]\mathbb P(E\cup F)=\dfrac{57}{100}+\dfrac{13}{20}-\dfrac{17}{50}=\dfrac{22}{25}[/tex]