Conditional Probability of a single event The question: Find the probability that a randomly selected person does not catch the 'flu' in terms of V and F. 


*

*V is the event that a person has been vaccinated

*F is the event that a person catches the flu.


Where 80% of population has been vaccinated against the flu, 


*

*but 5% of the vaccinated population catches the flu anyway.

*95% chance to not catch the flu


So 20% of the population that have not been vaccinated 


*

*have a 20% chance to catch the flu.

*have an 80% chance to not catch the flu


My attempt:
So according to the information provided by the question, I built a probability tree diagram and from that diagram I tried to express the answer as (F'|V) ∪ (F'|V'). Where (F'|V) = 0.76 and (F'|V') = 0.16. But I do not know if the way I have done it, is correct or not and if it is correct, I don't know how to further proceed. 

Actually you have $P(F' \mid V)=0.95$ and $P(V)=0.80$ so $P(F' \cap V)=0.76$
Similarly $P(F' \mid V')=0.80$ and $P(V')=0.20$ so $P(F' \cap V')=0.16$ 
Since these are exclusive events, $P(F')=P((F' \cap V) \cup (F' \cap V')) = 0.76+0.16=0.92$
