Question

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

Answer 1

By the inclusion/exclusion principle,

$\mathbb P(E\cup F)=\mathbb P(E)+\mathbb P(F)-\mathbb P(E\cap F)$

Recall the law of total probability:

$\mathbb P(F)=\mathbb P(E\cap F)+\mathbb P(E^C\cap F)$

We're given

$P(F^C)=\dfrac7{20}\implies\mathbb P(F)=1-\mathbb P(F^C)=\dfrac{13}{20}$

and so

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

Now,

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