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, $P(F')=P((F' \cap V) \cup (F' \cap V')) = 0.76+0.16=0.92$