. (25 points) Suppose the probability of exposure to the flu during an epidemic is 0.6. A vaccine is 80% successful in preventing an inoculated person from acquiring the flu, if exposed to it. A person not inoculated faces a probability of 0.9 of aquiring the flu, if exposed to it. Define the probability of at least one will be flue for a randomly selected person.

Answer:The probability of at least one person will catch the flu is 0.5952Step-by-step explanation:To find the probability of al least one catches the flu, we need to find the difference between 1 and the probability of not catching the flu.The probability of catching the flu varies if the person is inoculated or not.Let the events be:E: the person was exposedF: the person catches the fluHence, it is attached a tree diagram of the sample space. According to the diagram:Inoculated person:P(F')= 0.6×0.8 + 0.4×1 = 0.88 (We choose the second and the fourth way)Not inoculated person:P(F')= 0.6×0.1 + 0.4×1= 0.46 (We choose the second and the fourth way)Let the events be:[tex]F_{Inoculated}[/tex]= the person is inoculated and catches the flu[tex]F_{Not inoculated}[/tex]=the person is not inoculated and catches the fluUsing complementary events, DeMorgan’s Laws, and independence we have:P([tex]F_{Inoculated}[/tex]∪[tex]F_{Not inoculated}[/tex])=1-P([tex]F_{Inoculated}[/tex]∪[tex]F_{Not inoculated}[/tex])'=1 - P([tex]F'_{Inoculated}[/tex]∩[tex]F'_{Not inoculated}[/tex])=1 - P([tex]F'_{Inoculated}[/tex])×P([tex]F'_{Not inoculated}[/tex])=1 - (0.88)×(0.46)=1 - 0.4048=0.5952The probability of at least one will catch the flu for a randomly selected person is 0.5952